R. M. M. - 26...GHEORGHE CĂINICEANU-ROMANIA D.M. EDITORIAL BOARD EMILIA RĂDUCAN BĂTINEȚU-GIURGIU...
Transcript of R. M. M. - 26...GHEORGHE CĂINICEANU-ROMANIA D.M. EDITORIAL BOARD EMILIA RĂDUCAN BĂTINEȚU-GIURGIU...
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ROMANIAN MATHEMATICAL SOCIETYMehedinți Branch
AUTUMN EDITION 2020
R. M. M. - 26 ROMANIAN MATHEMATICAL
MAGAZINE
ISSN 2501-0099
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1 ROMANIAN MATHEMATICAL MAGAZINE NR. 26
ROMANIAN MATHEMATICAL
SOCIETY
Mehedinți Branch
ROMANIAN MATHEMATICAL MAGAZINE
R.M.M.
Nr.26-AUTUMN EDITION 2020
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ROMANIAN MATHEMATICAL
SOCIETY
Mehedinți Branch
DANIEL SITARU-ROMANIA EDITOR IN CHIEF
ROMANIAN MATHEMATICAL MAGAZINE-PAPER VARIANT ISSN 1584-4897
GHEORGHE CĂINICEANU-ROMANIA
EDITORIAL BOARD
D.M.BĂTINEȚU-GIURGIU-ROMANIA
CLAUDIA NĂNUȚI-ROMANIA
NECULAI STANCIU-ROMANIA
DAN NĂNUȚI-ROMANIA
IULIANA TRAȘCĂ-ROMANIA
EMILIA RĂDUCAN-ROMANIA
DRAGA TĂTUCU MARIANA-ROMANIA
DANA PAPONIU-ROMANIA
GIMOIU IULIANA-ROMANIA
DAN NEDEIANU-ROMANIA
OVIDIU TICUȘI-ROMANIA
MARIA UNGUREANU-ROMANIA
ROMANIAN MATHEMATICAL MAGAZINE-INTERACTIVE JOURNAL ISSN 2501-0099 WWW.SSMRMH.RO
DANIEL WISNIEWSKI-USA
EDITORIAL BOARD VALMIR KRASNICI-KOSOVO
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CONTENT
ABOUT SOME ION IONESCU’S INEQUALITIES – D.M. Bătinețu-Giurgiu, Daniel Sitaru,Neculai
Stanciu ................................................................................................................................... 4
A GENERALIZATION OF FINSLER-HADWIGER’S INEQUALITY - D.M. Bătinețu-Giurgiu, Daniel
Sitaru, Claudia Nănuți ............................................................................................................. 9
A SIMPLE PROOF OF J.RADON’S INEQUALITY(1913) - Daniel Sitaru .................................... 11
ABOUT CALCULUS OF SOME LIMITS - Florică Anastase ........................................................ 12
ABOUT NAGEL AND GERGONNE’S CEVIANS – Bogdan Fuștei ................................................. 16
SOLVED PROBLEMS - Florentin Vișescu ................................................................................ 23
SOLVING SOME PROBLEMS WITH DETERMINANTS - Marian Ursărescu............................... 28
STRUCTURI ALGEBRICE (III) – Vasile Buruiană ...................................................................... 31
ABOUT SOME CLASSES OF SEQUENCES – Florică Anastase .................................................. 35
ABOUT PROBLEM JP.207 - RMM-SPRING EDITION 2020 - Marin Chirciu .............................. 42
ABOUT PROBLEM JP.227 - RMM-SPRING EDITION 2020 - Marin Chirciu .............................. 45
ABOUT PROBLEM JP.228 - RMM-SPRING EDITION 2020 - Marin Chirciu .............................. 50
ABOUT PROBLEM JP.232 - RMM-SPRING EDITION 2020 - Marin Chirciu .............................. 54
ABOUT PROBLEM JP.247 - RMM-SUMMER EDITION 2020 - Marin Chirciu ........................... 57
PROPOSED PROBLEMS ......................................................................................................... 60
INDEX OF PROPOSERS AND SOLVERS RMM-26 PAPER MAGAZINE .................................... 103
REVIEW OF THE BOOK “CALCULUS MARATHON”………………………………………………………...104
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ABOUT SOME ION IONESCU’S INEQUALITIES
By D.M. Bătinețu-Giurgiu, Daniel Sitaru,Neculai Stanciu
If ( ), then the following inequalities hold:
(N-I)
namely NESBITT-IONESCU inequality, and also:
(*)
Let be a triangle, with the usual notations and is the semi-perimeter, and its area.
In Mathematical Gazette, 1897, ION IONESCU has established that, in any triangle the
following inequality holds: √ (I-W) and this inequality has been rediscovered by ROLAND WEITZENBOCK, in 1919. Also, in Mathematical Gazette, vol., XLVIII, (1942-1943) at page 334, it is proved that, in any triangle the following inequalities hold: ( ) (**) (***).
Next, we will establish other inequalities (some known):
TSINTSIFAS’S INEQUALITY:
If , then in any triangle the following inequality holds:
√ (T)
Proof. We have:
∑
∑
∑
( )
∑
( )
∑(
*
( ) ( ) ∑
( )
( ) ( )
∑ ( ) ( )
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( )
( ) ( ) ( )
( )
( ) ( )
( √ ) √ √
BĂTINEȚU-GIURGIU’S INEQUALITY:
If , then in triangle the following inequality holds:
√ (B-G)
Proof 1. We have:
∑
∑√
√∏√
√( )
√
√
Proof 2. We have:
∑
∑( )
∑
( )
√ √
Proof 3. We have:
∑
∑
∑
( )
∑(
*
( )
( ) ∑
( )
( ) ( )
( )
( ) ( )
√ √
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THE SECOND INEQUALITY OF G. TSINTSIFAS
If , then in any triangle the following inequality holds:
(G.T.)
Proof 1. We have:
∑
∑
∑( )
( )
∑ ( )
( )
( )
( )
( )
Proof 2. We have:
∑
∑(
*
( ) ( ) ∑
( )
( ) ( )
∑ ( ) ( )
( ) ( )
( ) ( )
( )
( )( )( )( )
( )( )( )
Theorem 1. If , then in any triangle the following inequality holds:
(i)
Proof. We have:
∑
∑
∑( )
( )
∑ ( )
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( )
( )( )
( )
( )( )
If in (i) we take we obtain inequality (G.T.)
BĂTINEȚU-GIURGIU-SITARU INEQUALITY
If , then in any triangle the following inequality holds:
(B-G-S)
Proof 1. We have:
∑
∑( )
∑
( )
Proof 2. We have:
∑
∑√
√∏√
√( )
.√( )
/
4 √
5
Proof 3. We have:
∑
∑(
*
( )
( )∑
( )
( ) ( )
( ) ( ) ( ) ( )
( )
( √ )
Theorem 2. If , then in any triangle the following inequality holds:
(ii)
Proof. We have:
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∑
∑( )
∑
( )
Theorem 3. If , then in any triangle the following inequality holds:
√ (iii)
Proof. We have:
∑
√ ∑√
√ √∏√
√ √( )
√ .√( )
/
√ 4 √
5
√
√
If relationship (ii) is transformed in relationship (B-G-S), and if relationship (iii), becomes relationship (B-G-S). If in (B-G-S) inequality we obtain inequality:
(F.G) namely inequality F. GOLDNER.
Theorem 4. If , then in any triangle the following inequality holds:
√ (j)
Proof 1. We have:
∑
∑( ) ( )
( ) ∑
( ) ( )∑
( )
( )( )
∑ ( ) ( )
( )
( ) ( ) ( )
( )
( ) ( )
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( √ ) ( )
√ ( √ ) √ √
Proof 2. According to inequalities (T) and (B-G) we have:
∑
∑
∑ ( )
∑
√ √ √ . We conclude with some remarkable observations
1. If in inequalities (T), (B-G), (j) we take , we obtain Ion Ionescu’s inequality, namely inequality (I-W).
2. If triangle is equilateral then, from inequalities (T), (G.T) we obtain inequality (N-I), and from inequalities (B-G), (B-G-S), we obtain inequality (*).
3. If triangle is equivalent then from inequality (j) we obtain inequality:
namely
Bibliography
[1] Bătinețu-Giurgiu M.D., Boroica Gh., Mușuroia N., About some inequalities of Ion Ionescu, Argument Magazine, Year 20, 2018, pp 12-17.
[2] Stoica Gh., A generalization of Tsintsifas inequality, Mathematical Magazine from Valea Jiului, Nr. 2, September, 2018, pp. 25-27.
[3]Romanian Mathematical Magazine-Interactive Journal-www.ssmrmh.ro
A GENERALIZATION OF FINSLER-HADWIGER’S INEQUALITY
By D.M. Bătinețu-Giurgiu, Daniel Sitaru,Claudia Nănuți
ABSTRACT: Finsler-Hadwiger’s inequality:
In any triangle having the area , the following inequality holds:
√ ( ) ( ) ( ) (F-H)
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Let’s suppose and generalize this inequality, but first, we state some important results.
IMPORTANT RESULTS:
In Mathematical Gazette vol. XLVIII (1942-1943) at page 334 it is proved that:- in any triangle the following inequality holds: ( ) (i)
and (ii)
where is the semiperimeter of the triangle. Also, we will take into account Doucet’s
inequality: √ (D) and D.S. Mitrinovic’s inequality: √ (M)
MAIN RESULT:
If ( ) 0
1, then in any triangle having the area has the
inequality: ( ) √ (( ) ( ) ( ) ) (*)
Proof. We have:
∑
∑( )
( )∑
∑
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( √ ) ( ) ( ) ( )
( )√ ( ) √ ( )√ √ Q.E.D.
If then inequality (*) becomes ( ) √ ∑ ( )
√ ( ) ( ) ( ) , namely inequality (F-H).
If
then inequality (*) becomes: ( ) √
∑ ( )
√
(( ) ( ) ( ) )
REFFERENCES:
Romanian Mathematical Magazine-Interactive Journal-www.ssmrmh.ro
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A SIMPLE PROOF OF J.RADON’S INEQUALITY(1913)
If * + then:
( )
( )
By Daniel Sitaru-Romania
Proof: Let be ( ) ( )
( )
( )
( )
( ) ( ( ))
( )
( ) 4 ( )
5 4
( )
5
4 ( )
5
( )
( )
( )
(
*
(
*
(
*
(
*
Analogous: ( )
.
/
.
/
( ) ( ) ( )
( ) (
*
(
*
( )
( )
( ) ( )
(
*
(
*
(
*
(
*
( )
( ) (
*
(
*
( )
( )
Analogous for : * + then:
( )
( )
( )
( )
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Equality holds for:
The reverse Radon’s inequality: If * + then:
( )
( )
Refferences: https://www.cut-the-knot.org/m/Algebra/RadonInequality.shtml
Romanian Mathematical Magazine-Interactive Journal-www.ssmrmh.ro
ABOUT CALCULUS OF SOME LIMITS
By Florică Anastase-Romania
Abstract: In this paper is given a method for calculus of some limits.
Main result:
If ( ) continuous and ( )
, it follows:
( ∑( 4
5+
+
Proof:
( )
( )
( ) ( ) ( )
( )
( ( ))
( )
( 4
5+
( )
( )∑
∑( 4
5+
( )∑
∑
∑(
*
∫
https://www.cut-the-knot.org/m/Algebra/RadonInequality.shtml
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(∑( 4
5+
+
Application 1:
For find:
( ∑ √ 4
5
)
Solution: By ( )
such that:
4 .
/5
(
*∑
∑ √ 4
5
(
*∑
∑
∑(
*
∫
( ∑ √ 4
5
)
Application 2: For find: 4 ∑ √ .
/
5
Solution: By ( )
such that:
. .
//
(
*∑
∑ √ 4
5
(
*∑
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∑
∑(
*
∫
( ∑ √ 4
5
)
Application 3: For find:
( ∑ √ 4
5
)
Solution: By ( ( ))
such that:
. .
//
(
*∑
∑ √ 4
5
(
*∑
∑
∑(
*
∫
( ∑ √ 4
5
)
Application 4: For find:
( ∑ √(
*
)
Solution: By ( )
such that:
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(
*
(
*∑
∑ √
(
*∑
∑
∑(
*
∫
( ∑ √
+
Application 5: For ( ) , find:
∑ 4
5
Solution:
∑ 4
5
∑( .
/
)
∑( .
/
)
such that: ( (
*
+
( )∑
( )∑
∑
∑(
*
∫
∑ 4
5
Application 6: For find:
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(
( ∑ √ 4
5
))
Solution: By ( )
such that:
(
*
√ 4
5
(
*
(
*∑
∑ √ 4
5
(
*∑
By ∑
∫
, we have:
4 ∑ √ .
/
5
, hence:
(
*
( ) (
*
.
/ .
/
Proposed problems:
)
∑(√
)
)
∑(√
)
Refferences: Romanian Mathematical Magazine-Interactive Journal - www.ssmrmh.ro
ABOUT NAGEL AND GERGONNE’S CEVIANS
By Bogdan Fuștei-Romania
ABSTRACT: In this Math Note we will establish a new geometric identity in triangle:
(and the analogs),
http://www.ssmrmh.ro/
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where, Nagel’s cevian from Gergonne’s cevian from being the usual notations. This identity was proposed as a problem for RMM Magazine (IDENTITY IN TRIANGLE 122). In this note we will propose the way we’ve obtained this identity and some of its applications.
( )
(and the analogs)
( )
( )
( )
( )
=( )
( )
( )( ) ( )
( ) ( )
( ) ( )
( )
(and the analogs) (
) ( ) (and the
analogs)
1. Let be (Nagel’s cevian from );
and using cosine theorem in we have:
( ) ( )
(and the analogs)
( ) 4
5 ( ) ( )
( )
( )4
5 ( )4
5
( )( )
4
( )( )
5
( ) 4 ( )( )
5
But ( )( ) ( ) (and the analogs) ( )
( )
(and the
analogs)
2. Using Stewart’s theorem, we obtain:
( ) ( ) ( )( )
(and the analogs)
( ) ( ) ( )( )
(and the analogs)
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( )
( )
( )
( ) ( )
(
)
3. Using the inequality between the squared mean and the arithmetic mean:
√
√
(and the analogs)
Which is equivalent with: √ √ (and the analogs), but (and
the analogs) √ √ (and the analogs).
But we know that
(and the analogs)
√
√ (and the
analogs). So, we’ll obtain the inequality:
√
√ (and the analogs)
Summing we will obtain a new inequality, namely:
√ ∑
√
∑
But ∑
∑
√ ∑
√
∑
But
and the analogs
√ ∑
√
∑
(and the analogs) ∑
√ ∑
√
4. We know that ( ) ∑
(
) ∑
∑ ∑
∑ ( ) ∑
∑
But
(and the analogs), so we will obtain a new identity, namely:
∑
∑
∑
5.
(Tereshin Inequality)
( ) (and the analogs)
( ) ∑( )
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.∑ ∑
/ ∑( ) √∑( )
∑ ∑
(and the analogs);
6. From the above expressions of and
we can easily observe that
(and the analogs)
So, we can write that:
∑
7. We know that
. So,
∑
∑
Taking into account the above we can write that:
. Summing we will
obtain the identity: ( ) ∑
( ) ∑
8. We know that ( )
√
(and the analogs)
( )
( )
( )
∑
∑ ∑
From the above we can write that: ∑(
) ∑ ∑( )
9. We know that
√
(and the analogs);
√
From the above we will have: √
(and the analogs) √
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From the above we will have: √
(and the analogs) √
√
Summing we will obtain: ( ) √
∑
√
Summing we will have the following inequality:
( )√
( ) ∑
10. From inequality
∑
( )
(Tereshin’s inequality)
.
/ ∑
∑
√ ∑
√
∑
√ ∑
√
∑
∑
11.
(Laurențiu Panaitopol –Mathematical Gazzete)
∑
√ ∑
√
So, finally, we have as being true a new inequality, namely:
∑
√ ∑
√
So,
( )
(and analogs);
(and analogs)
(and the analogs);
( )
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( )( ) (and the analogs)
12. We know that:
√
√
( ) (and the
analogs)
( ) ∑
( )
∑
∑
∑
, so, we have the following:
∑
(
*
From the above we have:
(and the analogs), but
So, we have the identity: ∑
( )
13. We proved that:
( ) (and the analogs).
But
(and the analogs) ( ) (and the analogs)
( ) (and the analogs)
(and the analogs)
Taking into account the above inequality we have:
∏( ) ∏( )
From the above we have: ∏(
)
But we will obtain the following: ∏(
)
∏(
)
we will obtain the following:
∏(
)
14. From
and
(and the analogs), we will obtain:
.
/ ( )
.
/
.
/
(and the analogs)
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.
/
.
/
.
/
.
/
(and the analogs)
( ) (
* ∑
∑
∑
So, finally we have the identity: ∑
.
/
Taking into account the above we have the following inequality:
(
* ∑
∑
(
* ∑
(and the analogs),
√
(and the analogs)
Squaring and taking into account:
(and the analogs)
( )
(and the analgos)
So, finally we have:
(and the analogs). Summing we will obtain: ∑
∑
15. But ∑
∑
∑
, so we will obtain the following: ∑
∑
∑
∑
.
/
(and the analogs);
√
(and the analogs)
∑(
*
∑
But ∏
so, we will have the inequality: ∏
16. From Tereshin’s inequality presented above and the new identity we will obtain the
following:
.
/
(and the analogs)
But
(and the analogs) and from the above we can write the following:
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(and the following)
we have the following:
(and the following)
Summing we will obtain a new inequality, namely: ∑
∑
.
/
(and the analogs)
(and the analogs)
√
(and the analogs). From the above we have:
(and the analogs)
Summing we will obtain: ∑
17. From
(and the analogs) we have ∑
∑
( ) ∑ ∑
∑
∑ ∑
These are just a little part of the relationships that can be obtained using the established relationship in this note. Any other new obtained relationships will be published in the form
of problems for RMM.
Refferences: Romanian Mathematical Magazine-Interactive Journal-www.ssmrmh.ro
SOLVED PROBLEMS
By Florentin Vișescu – Romania
1. Let ( ). We denote
We prove that:
a) If and then
b) If ( ) and
then
c) If √ and then
d) If ( √ ) and then
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Demonstration.
a), b) We consider the function ( )
( )
( ) ( )
( ) ( )
( )
( )
( )
( )
( )
( )
a) If ( ) convexe and then .
/
( ) ( ) ( )
or
.
/
as and
. Then
b) If ( ) concave and then .
/
( ) ( ) ( )
or
as ( ) ( ) and
Then:
c), d) We consider the function ( )
( )
( )
( )
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( ) ( ) ( ) ( )
( )
( √ )( √ )
( )
If √ ( ) convexe and then √ we have
.
/
( ) ( ) ( )
or taking into account that .
( )
namely
If ( √ ) ( ) concave and then ( √ ) we have:
.
/
( ) ( ) ( )
or, taking into account that
( )
namely
2. Let be ( ). We denote
( )( )( ) ( ) ( ) ( )
Prove that:
a) If then
b) If ( ) with
then
c) If with then
d) If then
Demonstration.
b), c) We consider the function ( )
( )
( )
( )
( )
( )
( )( )
( )
b) If ( ) ( ) concave .
/ ( ) ( ) ( )
( ) or
.
/
Taking into account that
we obtain
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.
/ or
namely
c) If ( ) convexe .
/ ( ) ( ) ( )
and taking into account that we obtain
.
/ or
namely .
a) We consider the function ( ) ( )
( )
( )
( )
( ) concave. Then .
/ ( ) ( ) ( ) as we choose
and we obtain .
/ ( ) ( ) ( )
as ( ) ( ) ( ) ( ) or
( ) [ ( )( )( )]
( ) ( )( )( ) or
d) We consider the function ( ) ( )
( ) ( )
( )
( )
( )
for . So, concave on ( ). Then we have:
.
/ ( ) ( ) ( ). But as ( ) ( ) ( ) ( )
or ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) namely
3. a) Let be ( ) ∫ ∫ ∫
√( )
and ( ) ( )
Prove that if then ( ) ( ) and if then ( ) ( )
b) Let be ( ) ∫ ∫ ∫
( )
and ( )
( )
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Prove that if √ then ( ) ( ) and if √
( ) ( )
c) Let be ( ) ∫ ∫ ∫
(.
/ *
and ( ) .
/
Prove that if then ( ) ( ) and if then ( ) ( )
d) Prove that if then ∫ ∫ ∫
. √( )
/
.
/
Demonstration
a) According to the demonstration from example 1. a) and b), we have:
√( )
for and
√( )
for
( ).
Applying the triple integral, we obtain the desired inequality.
b) According to the demonstration from example 1) c) and d) we have:
.
/
.
/ for (√ ) and
.
/
.
/ for ( √ )
Applying the triple integral, we obtain the inequalities.
c) According to the demonstration from example 2. b) and e), we have:
(.
/ *
for ( ) and
(.
/ *
for .
Applying the triple integral, we obtain inequality d) according to the demonstration from 2)
a), we have
. √( )
/
for . Applying the triple integral, we
obtain the inequality.
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SOLVING SOME PROBLEMS WITH DETERMINANTS
By Marian Ursărescu-Romania
Abstract: In this article, we will solve some problems with determinants. A lot of these problems had appeared in math magazines or were proposed to various mathematic contests. For start we will remember the next lemma:
Lemma: Let be ( ). Then, ( ) ( ) is a polynomial function having the grade , which has the form: ( )
Remarks:
1. If ( ), then , -; if ( ) then , -, and if ( ) then , -
2. If ( ) then:
( ) , where (
) or ( ) or
( )
Applications:
1.Let be ( ) such that and ( ) .
Prove that .
Proof: ( ) ( ) ( ) ( )
(we have used the fact that ) ( ) or ( ) (1)
Let be ( ) ( ) . From (1) ( ) or
( ) ( )
2. Let ( ) such that ( ) . Prove that ( )
(NMO-Romania)
Proof: Let be ( ) ( ( )) , -
( ) ( ) ( )
We have ( ) ( ) ( ( ))
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( ) ( ) ( ( ))
But and from hypothesis ( ) and because
( ) ( ) ( )
3. Let be ( ) such that . Prove that the matrix √ is invertible. (Mathematical Gazette)
Proof: We must prove that ( √ ) . By absurdum suppose that (
√ ) . Let ( ) ( )
, with (√ ) √ √ √ ( )
√
, false, √
4. Let ( ) such that and ( ) are odds. Prove that the matrix is invertible . (Mathematical Gazette)
Proof:We must prove that ( ) . By absurdum suppose that such that ( ) . Let be ( )
, -. We have: ( ) odd. ( ) ( ) odd.
But ( ) (from Bézout) ( ) ( ) ( ) , -
( ) ( )
( ) ( ) ( ) } are odd numbers, false, because are
consecutives.
5. Let ( ) and
.
Prove that: ( ) ( ) ( ) ( )
(RMO-Romania)
Proof:Let be ( ) ( )
( )
( )
( ) (
) ( )
By summing ( ) ( ) ( )
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( ) (
( ) )
( ( ) ) (1)
But is the root having the order of the unit (2)
But (1) (2) ( ) ( ) ( )
(( ) )
( )
In the ending, some proposed problems on R.M.M.:
1. Let ( ) with such that ( ) . Prove that:
( ) (RMO-Romania)
2. Let be ( ) such that ( ) ( ) . Prove that:
( ) ( ) (R.M.M.)
3. Let be ( ) such that ( ) . Prove that:
( ) ( ) (R.M.M.)
4. Let be ( ). Prove that:
( ) ( ) ( ) ( )
(R.M.M.)
5. Let be ( ) such that ( ) . Prove that:
( ) .
/ ( ) (R.M.M.)
6. Let be ( ) such that . Prove that:
( ) ( ) (R.M.M.)
7. Let ( ). Prove that: ( ) ( ) ( ( ))
8. Let be ( ). If odd, then the matrix is invertible,
(R.M.M.)
Refferences: Romanian Mathematical Magazine-Interactive Journal - www.ssmrmh.ro
http://www.ssmrmh.ro/
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STRUCTURI ALGEBRICE (III)
By Vasile Buruiană – Romania
Exemple si observații
1. este factorial, deoarece cu ( ) | | arată că ( ) este inel Krull cu toți divizorii principali.
2. Orice inel Krull pentru care are un număr finit de atomi este factorial.
Într-adevăr: fie
un divizor și descompunerea sa unică în atomi. Există
astfel că ( ) și ( ) cu
unitatea lui , deci ( ) și deci orice divizor este principal.
În particular corpurile comutative sunt inele factoriale.
3. Inelul , - nu este inel Krull (deci nici factorial) fiindcă nu îndeplinește condiția h) din teorema 1 pentru inelele Krull: inelului, dar inelului.
4. Orice inel principal este factorial.
Demonstrație. PAS 1. Pentru orice șir de ideale ( ) ( ) astfel încât
( ) ( )
Într-adevăr luăm ( ) care este principal, deci cu ( ) astfel ca
( ) ( ) ( ) ( ) ( ) ( ) ( )
PAS 2. ( ) orice ireductibil este prim). Într-adevăr există astfel ca și evident ( )
PAS 3. inversabil există un divizor ireductibil al său. Într-adevăr dar este ireductibil, afirmația este adevărată. Dacă este ireductibil neunități și neasociate cu astfel ca . Dacă sau sunt ireductibile afirmația este dovedită. Astfel există
astfel ca , etc ( )
( )
( ) ceea ce nu este posibil la infinit
există un pas astfel ca | și este ireductibil.
PAS 4. Orice se descompune în produs finit de elemente ireductibile. Într-adevăr prin PAS 3, ex | ireductibil cu . Pentru , continuăm, ș.a.m.d. Prin pasul procesul se oprește într-un număr finit de pași.
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PAS 5. Cu PAS 2 și PAS 4 folosind Teorema 2.e. rezultă afirmația. Cum orice inel euclidian este principal orice inel euclidian este și el factorial.
5. Teorema 3 (Gauss): Dacă este inel factorial atunci , - este inel factorial.
Demonstrația se face în mai mulți pași.
PAS 1. este inel integru , - este integru în care unitățile sunt cele din (deci în , - unitate cu și , - atunci | | coeficienții lui )
Într-adevăr
cu și
{
PAS 2. Dacă este prim din este prin în , -. Într-adevăr dacă | și și coeficienți ai astfel ca . Fie și minim cu proprietatea că
. Deoarece | ∑
∑
și |∑
| | sau | ceea ce este
absurd.
PAS 3. Pentru , - notăm (( ) ( ) și îl numim
conținutul lui cu (( ) se numește primitiv.
Avem (( ) (( ) ( ))
Într-adevăr fie ( ) ( ) ( ) ( ) deci ar fi suficient să arătăm că produsul a două polinoame primitive este primitiv. Dar acesta nu ar fi adevărat ci rezultă
că , prim, cu |
| sau |
sau nu este primitiv, ceea ce este absurd.
PAS 4. Pentru , - primitiv și | | (dacă și este primitiv și
cu )
Într-adevăr | , - cu ( ) ( ) ( )
( ) ( ) căci este primitiv |
PAS 5. Dacă grad pentru , -, atunci sunt echivalente
a) este ireductibil în , -
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b) este primitiv și ireductibil în , -, unde este corpul de fracții al lui .
Demonstrație.
a) b) Evident este primitiv. Dar ar fi reductibil în , - , - , grad , grad grad și grad , grad cu unde au grade egale cu ale respectiv ( îl putem lua produsul numitorilor lui și ). Dacă ( ) ,
( ) cu primitiv
| deci este reductibil și , -, ceea ce este contradictoriu.
b) este evidentă.
PAS 6. ireductibil în , - este prim.
Într-adevăr, pentru grad , afirmația este evidentă pe baza factorialității lui și pas 2.
Dacă grad și | este prim în , - | sau | în , -. Dacă | și
în astfel ca , - | în , -
| în , -
Analog dacă | este prim.
PAS 7. Orice element din , - este produs finit de elemente prime. Este suficient să aratăm că orice element nenul și neinversabil din , - este produs finit de elemente ireductibile.
Într-adevăr pentru grad afirmația este evidentă.
Dar grad fie ( ) cu primitiv. Dacă este ireductibil, afirmația este demonstrată. Dar cu având grade mai mici decât al lui , atunci prin reducție ele se descompun în produse finite și deci și .
PAS 8. Pe baza PAS 7 și PAS 6 , - este factorial.
6. Observație. În pasul al sașelea am folosit faptul că inelul , - este factorial. Aceasta rezultă independent de demonstrația precedentă, aratând că este inel euclidian. Într-adevăr, pentru , -, există , - astfel încât și ori ori
grad grad . Într-adevăr fie
. Folosind inducție după avem: Pentru și afirmația este evidentă.
Presupunem că ea este divizată pentru și fie de gradul
are grad
cu și ori ori grad grad
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.
/ deci luând
și
obținem afirmația și pentru . Luând acum , - * + prin ( ) grad , obținem euclidianeitatea (deci și factorialitatea) lui , -
7. În general, dacă este factorial , - este factorial după cum rezultă folosind exemplul 6. Aplicând metoda reducție.
8. Există inele factoriale neprincipale. Într-adevăr: fie inel factorial care nu este corp neinversabil. Fie în , - idealul ( ). Dacă acest ideal ar fi principal, ar exista , - astfel ca , - , - , - | și | și este inversabil , - , - , - astfel ca ( ) ( ) . Nu se poate ca căci ar rezulta inversabil . Facând ( ) deci iarăși o absurditate
nu avem că ( ) este principal. Deoarece am văzut că , - este factorial , - este factorial și nu este principal când este inel factorial care nu este corp.
9. , - * | + (inelul întregilor lui Gauss) este inel euclidian deci este factorial (și principal).
Demonstrație. Definim ( ) ( )( ) numită norma lui
. Evident ( ) ( ) ( ). Apoi , -, există , - astfel ca
și ori ori ( ) ( ). Într-adevăr dacă
, - cu
astfel ca | |
| |
. Luăm .
Avem .
/ ( ) ( )
și ( ) 0 .
/1 ( ) .
/
( ) ( ). Deci ( ) ( ) verifică faptul că , - este inel euclidian.
10. Fie [ √ ] { √ | } care este inel integru. El nu este factorial. Într-adevăr
fie ( √ ) ( √ )( √ ) (norma)
Evident ( ) ( ) ( ) și | ( )| ( ). Apoi ( este unitate în [ √ ])
( ( ) ) * +. În fine este un element ireductibil care nu este prim. Într-adevăr, dacă ( ) ( ) ( ) ( ) ( ). Cum se consideră neunități și neasociate cu ( ) ( ) . Cum nu are soluții în
nu există neunități și neasociate cu astfel că este ireductibil în [ √ ].
Nu este prim însă fiindcă | ( √ )( √ ) și 3 √ căci dacă | √
( )| ( √ ) | ceea ce este fals.
11. Inelul , - de asemenea nu este factorial fiindcă nu există ( ).
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12. Dacă este factorial și și ( ) și deci cu (identitatea lui Bezout)
13. Fie inel integru în care orice element nenul și neinversabil se descompune în produs finit de elemente ireductibile.
Atunci: Acest factorial , idealul (în acest caz, , -)
Demonstrație:
” ” Dacă este factorial , -. Avem caci dacă
| , iar dacă
deci . Reciproc: , -
( ) deci cu Teorema 2 avem că este factorial.
ABOUT SOME CLASSES OF SEQUENCES
By Florică Anastase-Romania
ABSTRACT: In this article are studied few classes of sequences gived by initial conditions and a nonlinear formula of recurrence.
Application 1: ( ) is a sequence of real numbers with
√
. Find:
( )
Solution: From
( ) descending and ( ) is convergent, then
.
/ ( )
( )
√ .
/
( )
Application 2: ( ) is a sequence of real numbers ( ), and √
. Find:
( )
Solution: We prove: ( ) √
√ √
√
√ (√ √ )
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( )( )
√ (√ √ ) ( ) is increasing. So: ( ) is convergent and
, .
/ for and
√
( )
.
/
[.
/
]
.
/
.
/
Application 3:
If ( ) is a sequence of real numbers ( ), and √ √
. Find:
( )
Solution: We prove to: ( ) , and ( ) then .
/
for which . We have:
√ √
. So:
( )
Application 4:
( ) is a sequence of real numbers and √ . Find:
Solution: √ ( ) is increasing and
( ) sequence of real numbers .
/
√
√ .
/
.
/
.
/
is a geometric progression with:
and
.
/
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.
/
.
/
.
/
Application 5: ( ) is a sequence of real numbers
√
√
Find:
Solution:
√
and
√
√
√
.
/ √
√
⏟
.
/
Application 6: Find:
(
√
√
( √
) √
( √
( √
),
⏟ )
Solution:
Let
(
√
√
4 √
5 √
( √
4 √
5)
⏟ )
√
√
( √
)
√
( √
( √
),
⏟
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then:
Application 7: ( )
.
/ ( ) is a sequcen of real numbers and
.
/. Study the convergence of sequence ( ) and find: ( )
Solution: From
∏
∏
.
/
From
, and
.
/ then
( ) ( ) is convergent.
.
/
(
+
(
+
Refferences: Romanian Mathematical Magazine-Interactive Journal- www.ssmrmh.ro
ABOUT PROBLEM JP.207-RMM-SPRING EDITION 2020
By Marin Chirciu – Romania
1) In the following relationship holds:
Proposed by George Apostolopoulos – Messolonghi – Greece
Solution: Lemma:
1) In the following relationship holds:
( )
http://www.ssmrmh.ro/
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Proof. We use
hence:
∑
∑
∑ ( )( )
( ), which is a consequence of:
∑ ( )( ) ( )
Back to the main problem: By Lemma the inequality becomes: ( )
,
which is equivalent with Euler’s , ( )( ) and . Equality holds for an equilateral triangle.
Remark. Replacing by we obtain:
2) In the following relationship holds:
Proposed by Marin Chirciu – Romania
Solution: Lemma
3) In the following relationship holds:
Proof: By
we obtain:
∑
∑
∑
∑
( )
. Which is a consequence of: ∑
( )
Back to the main problem: By Lemma inequality can be written:
For the right-hand inequality:
( ) ( )
Case 1). If ( ) , the inequality is obvious.
Case 2). If ( ) , the inequality can be written:
( ) ( )
Which follows by Gerretsen’s inequality: . It remains to prove that:
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( ) ( )( )
( )( ) , obviously by Euler .
For the left-hand inequality:
. Which is a
consequence of Gerretsen’s inequality: . It remains to prove:
(Euler’s inequality)
Equality holds for an equilateral triangle.
Remark. Between sums
and
. The following
relationship exists:
4) In the following relationship holds:
Proposed by Marin Chirciu – Romania
Solution: By lemmas ∑
and ∑
( ), the inequality can
be written:
( ) ,
Which follows by Gerretsen’s inequality: . It remains to prove that:
( )( )
Obviously by Euler’s inequality: . Equality holds for an equilateral triangle.
Refferences: Romanian Mathematical Magazine – Interactive Journal – www.ssmrmh.ro
ABOUT PROBLEM JP.227-16 RMM – SPRING EDITION 2020
By Marin Chirciu – Romania
1) Prove that in any triangle the following relationship holds:
(
) ( )
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
http://www.ssmrmh.ro/
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Solution: We use Hölder’s inequality:
( )
( ) , with
equality if and only if
and the known identities in triangle:
. We obtain:
( )
(
*
( )
( ) , wherefrom (
) ( ) . Equality if and only if the
triangle is equilateral.
Remark. Analogous we can propose:
2) In the following relationship holds:
( )
Solution: We use Bergström’s inequality:
( )
, with
equality if and only if
and the known identities in triangle:
. We obtain:
( )
( )
( ) , wherefrom
( )
Equality holds if and only if the triangle is equilateral.
Remark. The problem can be developed:
3) In the following relationship holds:
(
) ( ) , where
Proposed by Marin Chirciu – Romania
Solution: We use Hölder’s inequality:
( )
( ) and
, with equality if and only if
and the know identities in triangle
. We obtain:
( )
(
*
( )
( ) ,
wherefrom (
) ( )
Equality holds if and only if the triangle is equilateral.
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Remark: Replacing with we propose:
4) Prove that in any triangle the following relationship holds:
(
) ( )
Solution: We use Hölder’s inequality:
( )
( ) , with
equality if and only if
and the known identities in triangle.
. We obtain:
( )
(
*
( )
( ) , wherefrom
(
) ( ) . Equality holds if and only if the triangle is equilateral.
Remark.
5) In the following relationship holds:
( )
Solution: We use Bergström’s inequality
( )
, with
equality if and only if the
and the known identities in triangle
. We obtain:
( )
( )
( ) , wherefrom
( ) .
Equality holds if and only if the triangle is equilateral.
6) In the following relationship holds:
(
) ( ) , where
Proposed by Marin Chirciu – Romania
Solution: We use Hölder’s inequality
( )
( ) and
with equality if and only if
and the known identities in triangle
we obtain:
( )
(
*
( )
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( ) , wherefrom (
) ( )
Equality holds if and only if the triangle is equilateral.
Remark.
7) In the following inequality holds:
.
/ ( ) , where
Proposed by Marin Chirciu – Romania
Solution: We use Hölder’s inequality
( )
( ) and
, with equality if and only if
and the known identies in triangle
We obtain:
( )
.
/
( )
( ) , wherefrom (
) ( )
Equality holds if and only if the triangle is equilateral.
8) In the following relationship holds:
.
/ ( ) , where
Proposed by Marin Chirciu – Romania
Solution: We use Hölder’s inequality
( )
( ) and
with equality if and only if
and the known identies in triangle
We obtain:
( ) , where (
) ( )
Equality holds if and only if the triangle is equilateral.
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( )
.
/
( )
( ) , wherefrom
(
* ( )
Equality holds if and only if the triangle is equilateral.
9) In the following relationship holds:
(
) ( ) , where
Proposed by Marin Chiricu – Romania
Solution: We use Hölder’s inequality:
( )
( ) and
, with equality if and only if
and the known relationships in triangle
. We obtain:
( )
.
/
( )
( ) , wherefrom (
) ( )
Equality holds if and only if the triangle is equilateral.
10) In the following relationship holds:
(
) ( ) , where
Proposed by Marin Chirciu – Romania
Solution: We use Hölder’s inequality
( )
( ) and
, with equality if and only if
and the known relationships in triangle
We obtain:
( )
.
/
( )
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( ) , wherefrom (
) ( )
Equality holds if and only if the triangle is equilateral.
ABOUT PROBLEM JP.228-16 RMM SPRING EDITION 2020
Proposed by Marin Chirciu – Romania
1) Prove that if are the lengths of the sides of a triangle, then:
√
√
√
( )
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
Solution: We prove the stronger inequality:
2) In the following inequality holds: √
√
√
√
Solution: We have: (∑√
*
∑
∑√
√
∑
∑√
( )( )
∑
, which follows from:
∑
( )
√
( )( )
and ∑
, true from:
∑
∑
, because ∑
(from Jensen)
Equality holds if and only if the triangle is equilateral.
Remark: Inequality 2) is stronger than inequality 1):
3) In the following relationship holds: √
( )
Solution
√
( )
( )
( )
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( )
( )
( ), ( ) ( ) -
, ( ) ( )( )- ( ) ( )
We distinguish the following cases:
Case 1). If ( ) , obvious inequality.
Case 2). If ( ) , the inequality can be rewritten:
( ) ( ) , ( ) ( )( )-, which follows from
Blundon-Gerretsen’s inequality: ( )
( )
It remains to prove that: ( ) ( )
( )
( ),( )( ) ( )( )-
( )( )
Obviously from Euler’s inequality . Equality holds if and only if the triangle is equilateral.
Remark: The inequalities can be written:
4) In the following inequality holds:
√
√
√
√
( )
Solution: See inequalities 2) and 3).Equality holds if and only if the triangle is equilateral.
Remark: Let’s find an inequality having an opposite sense:
5) In the following relationship holds:
√
√
√
√
Solution: We have: (∑√
*
∑
∑√
√
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∑
∑√
( )( )
∑
( )
which follows from: ∑
( )
√
( )( )
and ∑
( )
, true from CBS:
We obtain: 4∑
5
(∑
√( )( )
)
(∑√
√( )( )*
∑ ∑
( )( )
( )
( )
We’ve used: ∑ ∑
( )( )
and
(Gerretsen). It follows that ∑
( )
Equality holds if and only if the triangle is equilateral.
Remark: We can write the double inequality:
6) In the following relationship holds:
√
√
√
√
√
Proposed by Marin Chirciu – Romania
Solution: See inequalities 2) and 5). Equality holds if and only if the triangle is equilateral.
Remark. The sequence of inequalities can be written:
7) In the following relationship holds:
( )
√
√
√
√
√
√
Solution: See inequalities 4), 6), Euler’s inequality and ( ) ( ). Equality holds if and only if the triangle is equilateral.
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ABOUT PROBLEM JP.232-16 RMM SPRING 2020
Proposed by Marin Chirciu – Romania
1) Prove that in any triangle the following relationship holds:
√
√
√
√
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
Solution: We prove the following: Lemma:
2) In the following relationship holds: √
√
√
√
Solution: We have: (√
√
√
*
∑
∑√
√
∑
∑
[.
/
]
.
/
, which follows from ∑
[.
/
] and
∑
. Let’s get to the main problem: Using the Lemma, the inequality can be written:
√
√
√ (Doucet’s inequality). Equality holds if and only if the
triangle is equilateral. Remark. Let’s find an inequality having an opposite sense:
3) In the following relationship holds: √
√
√
√
Proposed by Marin Chirciu – Romania
Solution: Using the Lemma, the inequality can be written:
√
√
√
(
*
4 √
5
( ) , which follows from Gerretsen’s inequality
( )
. It remains to prove that:
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( )
( ) ( )( )
Obviously from Euler’s inequality . Equality holds if and only if the triangle is equilateral.
Remark. We can write the double inequality:
4) In the following relationship holds: √
√
√
√
√
Solution: See inequalities 1) and 3). Equality holds if and only if the triangle is equilateral.
Remark. Replacing with we propose:
5) In the following relationship holds: √
√
√
√
Proposed by Marin Chirciu – Romania
Solution: We prove the following Lemma:
6) In the following relationship holds:
(√
√
√
)
( ) ( )
Solution: We have (√
√
√
*
∑
∑√
√
∑
∑
( ) ( )
( ) ( )
Which follows from ∑
( ) ( )
and ∑
Let’s get back to the main problem: Using the Lemma, the inequality can be written:
( ) ( )
( ) ( )
We distinguish the following cases:
Case 1). If ( ) , the inequality is obvious.
Case 2). If ( ) , the inequality can be written:
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( ) ( ), which follows from Blundon-Gerretsen’s inequality:
( )
( ). It remains to prove that:
( ) ( )
( )( ) (Euler’s inequality)
Equality holds if and only if the triangle is equilateral.
Remark. Let’s find an inequality having an opposite sense:
7) In the following inequality follows:
√
√
√
√
Proposed by Marin Chirciu – Romania
Solution: Using the Lemma, the inequality can be written:
( ) ( )
( ) ( ) which follows from Gerretsen’s inequality
( )
. It remains to prove that:
( )
( ( )) ( )
( )( )
Obviously from Euler’s inequality . Equality holds if and only if the triangle is equilateral.
Remark. The double inequality can be written:
8) In the following relationship holds:
√
√
√
√
√
Proposed by Marin Chirciu – Romania
Solution: See inequalities 5) and 7). Equality holds if and only if the triangle is equilateral.
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ABOUT PROBLEM JP.247-17 RMM SUMMER EDITION 2020
By Marin Chirciu – Romania
1) If then: (√ √ √ ) ( )
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
Solution: Lemma:
2) If then: ∑ ( ) ∑ ( )
Proof: It follows by Schur’s inequality:
( )( ) ( )( ) ( )( ) ,
for . Equality holds if and only if . Back to the problem:
Denote √ √ √ . The inequality can be written:
( ) ( ). By homogeneity becomes:
( ) ( ) ( )
∑ ∑ ∑ ∑ (1)
By (1) and Schur’s lemma it suffices to prove:
∑ ( ) ( ) ∑ ∑ ∑
∑ ( ) ∑ ( ) ∑ ∑ ( ) ∑
∑ ( ) ∑ ( ) , equality holds for .
Equality holds if and only if
.
Remark. The inequality can be generalized:
3) If then: (√ √ √ ) ( ) ( )( )
Proposed by Marin Chirciu – Romania
Solution: Denote √ √ √ . The inequality can be written:
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( ) ( ) ( )( )
By homogeneity it becomes:
( ) ( ) ( ) ( )( )
( )∑ ∑ ∑ ( )∑ (2)
By (2) and Schur’s lemma it suffices to prove:
( ) 0∑ ( ) ( )1 ∑ ∑ ( )∑
( )∑ ( ) ( ) ( ) ∑ ( )
( )∑
( )∑ ( ) ( ) ( ) ( )∑
( )∑ ( ) ( ) ( )
( )∑ ( )∑
( )∑ ( ) ( ) ( ) ( )∑
( )∑ ( ) ( )4∑ ( )5
Obviously, because ∑ ( ) ∑ ( )
Equality holds for . Equality holds if and only if
.
Note: For we obtain problem JP.247 – 17 RMM Summer Edition 2020, proposed by Nguyen Viet Hung, Hanoi, Vietnam. Application in triangle:
4) In if then:
(√
√
√
)
∑
Proposed by Marin Chirciu – Romania
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Solution: Replace in 3) ( ) by .
/ and using the relationship:
∑
.Equality holds if and only if .
References: Romanian Mathematical Magazine – Interactive Journal – www.ssmrmh.ro
PROPOSED PROBLEMS
5-CLASS-STANDARD
V.1. a) Let be the number ⏟
. Find the reminder of the division of the number
by .
b) Does number ⏟
divides with ?
Proposed Corina Maria Viespescu, Mihai Ionescu –Romania
V.2. Fie . Dacă ( ) și ( ) arătați că
. Proposed by Petre Stângescu-Romania
V.3. Aflați dacă restul împărțirii lui la este , iar diferența dintre câtul împărțirii
lui la și câtul împărțirii lui la este . Proposed by Petre Stângescu-Romania
V.4. Fie cu și . Arătați că .
Proposed by Petre Stângescu-Romania
V.5. Aflați numărul știind că la împărțirea cu dă restul , iar la împărțirea cu dă
câtul . Proposed by Petre Stângescu-Romania
V.6. Fie , cu și . Aflați și dacă câtul
împărțirii lui la este iar câtul împărțirii lui la este 30.
Proposed by Petre Stângescu-Romania
V.7. Aflați numărul de trei cifre ̅̅ ̅̅ ̅ pentru care câtul împărțirii lui ̅̅ ̅̅ ̅ la inversul său
este , iar restul este . Proposed by Petre Stângescu-Romania
http://www.ssmrmh.ro/
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V.8. Fie . Aflați restul împărțirii lui la .
Proposed by Petre Stângescu-Romania
V.9. Fie mulțimea * | |( )+. Aflați card ( ).
Proposed by Petre Stângescu-Romania
V.10. Aflați dacă . Proposed by Petre Stângescu-Romania
V.11. Fie cu | . Aflați dacă ( ) ( ) .
Proposed by Petre Stângescu-Romania
V.12. Find all prime numbers, such that: are prime numbers.
Proposed by Gheorghe Calafeteanu – Romania
V.13. If is a prime number prove that: ( ) is perfect square.
Proposed by Gheorghe Calafeteanu – Romania
V.14.
If each letter represents distinct non-zero digits then prove that there exist no solution.
Proposed by Naren Bhandari – Nepal
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical
Magazine-Interactive Journal. 6-CLASS-STANDARD
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VI.1. Se consideră dreptunghic în . Fie punctele ( ) cu este bisectoarea din și este mijlocul segmentului , -. Arătați că dacă avem:
1) ( )
2) este bisectoarea . Proposed by Petre Stângescu-Romania
VI.2. Determinați mulțimea *( ) | +.
Proposed by Petre Stângescu-Romania
VI.3. Fie astfel încât |( ). Să se arate că nu poate fi poate fi pătrat perfect. Proposed by Petre Stângescu-Romania
VI.4. Aflați toate numerele ̅̅ ̅̅ ̅ de trei cifre nenule pentru care
̅̅ ̅̅ ̅ ( ̅̅ ̅ ̅̅ ̅ ̅̅ ̅) . Proposed by Petre Stângescu-Romania
VI.5. Aflați numărul ̅̅ ̅̅ ̅̅ ̅ de patru cifre nenule știind că ̅̅ ̅̅ ̅̅ ̅ ̅̅ ̅̅ ̅ ̅̅ ̅̅ ̅ ̅̅ ̅̅ ̅ ̅̅ ̅̅ ̅
Proposed by Petre Stângescu-Romania
VI.6. Să se demonstreze că pentru orice și număr prim, numărul nu poate fi cub perfect. Proposed by Petre Stângescu-Romania
VI.7. Aflați numărul ̅̅ ̅̅ ̅̅ ̅ știind că ̅̅ ̅̅ ̅̅ ̅ este pătrat perfect.
Proposed by Petre Stângescu-Romania
VI.8. Suma a două numere naturale nenule este . Împărțind pe unul la celălalt obținem restul . Aflați numerele. Proposed by Petre Stângescu-Romania
VI.9. Fie oarecare. Să se arate că , -.
Proposed by Petre Stângescu-Romania
VI.10. Fie . Să se arate că: ( ) , -
Proposed by Petre Stângescu-Romania
VI.11. Fie . Demonstrați că ( ) , - dacă și numai dacă și sunt direct sau invers proporționale cu numerele respectiv .
Proposed by Petre Stângescu-Romania
VI.12. Un călător a parcurs un drum în trei zile. În prima zi a parcurs din cât a parcurs în
celelalte două zile. În a doua zi a parcurs din cât a parcurs în celelalte două zile. În ultima zi a parcurs km, număr prim. Știind că în fiecare zi numărul kilometrilor parcurși este un număr natural, aflați lungimea drumului parcurs de călător.
Proposed by Petre Stângescu-Romania
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VI.13. Aflați restul împărțirii la al numărului .
Proposed by Petre Stângescu-Romania
VI.14. Fie cu și ( ). Să se arate că | | .
Proposed by Petre Stângescu-Romania
VI.15. If prove that is divisible with if and only if is divisible
with 23. Proposed by Gheorghe Calafeteanu – Romania
VI.16. Prove that for the number can’t be a perfect square.
Proposed by Gheorghe Calafeteanu – Romania
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical Magazine-Interactive Journal.
7-CLASS-STANDARD
VII.1. Let be triangle, the middle of ( ) ( ) such that and
( ) . Prove that the points are collinears.
Proposed by Petre Rău – Romania
VII.2. Prove that for any two natural real numbers and and any nonzero natural number
the following inequality is true: ( ) ( ).
Proposed by Petre Rău –Romania
VII.3. Fie dreptunghic în . Dacă și ( ( )) sunt bisectoarea, respectiv
mediana din , iar este aria ( ), arătați că: √ .
Proposed by Petre Stângescu-Romania
VII.4 Fie cu ( ) sunt bisectoarea, respectiv înălțimea din
( ( ))
i) Arătați că
√ . ii) Dacă
√
, aflați raportul catetelor
Proposed by Petre Stângescu-Romania
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VII.5. Fie cu și . Să se arate că
.
Proposed by Petre Stângescu-Romania
VII.6. Să se demonstreze că nu există cu .
Proposed by Petre Stângescu-Romania
VII.7. În dreptunghic în , punctele ( ) sunt picioarele înălțimii
bisectoarei și medianei din . Dacă
, aflați măsurile unghiurilor ascuțite ale .
Proposed by Petre Stângescu-Romania
VII.8. În ( ). Dacă , arătați că
este dreptunghic. Proposed by Petre Stângescu-Romania
VII.9. În cu ( ). Dacă
, arătați că este
dreptunghic.
Proposed by Petre Stângescu-Romania
VII.10. Fie dreptunghic în . Se consideră ( ) astfel încât
. Aflați
( ). Proposed by Petre Stângescu-Romania
VII.11 Fie cu ( ) . În exteriorul se consideră semicercul de
diametru , -. Dacă să se arate că . ⏜/ .
Proposed by Petre Stângescu-Romania
VII.12. În ( ) . Știind că ( ) (√ ), aflați ( ) și
( ). Proposed by Petre Stângescu-Romania
VII.13. Fie cu și . Demonstrați că ( )
Proposed by Petre Stângescu-Romania
VII.14. Fie cu . Arătați că dacă astfel încât și
, atunci . Proposed by Petre Stângescu-Romania
VII.15. Dacă astfel încât și , arătați că atunci
. Proposed by Petre Stângescu-Romania
VII.16. Fie √ √ √ √ , unde cu . Aflați și știind că
media aritmetică și media geometrică a numerelor și sunt raționale.
Proposed by Petre Stângescu-Romania
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VII.17. For prove that: √
Proposed by Gheorghe Calafeteanu – Romania
VII.18. Find: 0√ √ √ √
√ √ 1 , - - great integer function.
Proposed by Gheorghe Calafeteanu – Romania
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical
Magazine-Interactive Journal. 8-CLASS-STANDARD
VIII.1. If then: * + 2
3 2
3 √* +
* + , - , - - great integer function
Proposed by Daniel Sitaru,Emil Stanciu – Romania
VIII.2. If ( ), then:
.
/
(
*
.
/
(
*
4.
/
(
*
.
/
(
*
5
4.
/
(
*
.
/
(
*
5
Proposed by D. M. Bătinețu – Giurgiu, Eugenia Turcu – Romania
VIII.3. If ( ), then:
.
/
(
*
.
/
(
*
Proposed by D. M. Bătinețu – Giurgiu, Lucian Lazăr – Romania
VIII.4. If , then:
4
54
54
5
( )( )( )
Proposed by D. M. Bătinețu – Giurgiu, Elena Nicu– Romania
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VIII.5. If , then:
(
*
Proposed by D. M. Bătinețu – Giurgiu, Daniela Micu – Romania
VIII.6. If ( ), then: .
/
.
/
.
/
Proposed by D. M. Bătinețu – Giurgiu,Sorin Pîrlea – Romania
VIII.7. 1. If we have the real numbers with and , prove that:
( )
( )
( )
Proposed by Petre Rău– Romania
VIII.8. Solve in the following system of equations: {( )( )
( )( )
Proposed by Mioara Mihaela Mirea, Sorin Dumitrescu– Romania
VIII.9. Aflați pentru care
.
Proposed by Petre Stângescu-Romania
VIII.10. Fie cifre nenule, și √ ⏟
̅̅ ̅̅ ̅̅ ̅̅ ̅̅ √ ⏟
̅̅ ̅̅ ̅̅ ̅̅ ̅̅ .
Aflați știind că . Proposed by Petre Stângescu-Romania
VIII.11. Să se arate că dacă și sunt cifre, și , atunci √ ̅̅ ̅̅ ̅̅ ̅
Proposed by Petre Stângescu-Romania
VIII.12. Fie cu . Să se arate că √ ( )
Proposed by Petre Stângescu-Romania
VIII.13. Fie dreptunghi în . Dacă este lungimea bisectoarei din și este
lungimea medianei din , aflați unghiurile ascuțite ale dacă
√ √
Proposed by Petre Stângescu-Romania
VIII.14. Aflați dacă .
Proposed by Petre Stângescu-Romania
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VIII.15. Un trapez ( ) cm, cm este în același timp înscris într-un cerc de rază și circumscris unui cerc de rază . Aflați și .
Proposed by Petre Stângescu-Romania
VIII.16. If then:
Proposed by Rahim Shahbazov- Azerbaijan
VIII.17. Solve for natural numbers: {
( )
Proposed by Gheorghe Calafeteanu – Romania
VIII.18. Solve for real numbers: ,
,
Proposed by Gheorghe Calafeteanu – Romania
VIII.19. Solve for real numbers: {
{
Proposed by Gheorghe Calafeteanu – Romania
VIII.20. Solve for integers:
Proposed by Gheorghe Calafeteanu – Romania
VIII.21. Solve for real numbers: {
Proposed by Gheorghe Calafeteanu – Romania
VIII.22. Let such that . Prove that:
(
*
Proposed by Marin Chirciu – Romania
VIII.23. Let and . Prove that:
Proposed by Marin Chirciu – Romania
VIII.24. If denotes the distinct or identical digits and
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ie there exist the multiple of 17 such that sum of the digits is . Find such smallest and largest four-digit multiple of . Proposed by Naren Bhandari – Nepal
VIII.25. If then:
( )( )( )
(
*
Proposed by Daniel Sitaru-Romania
All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical
Magazine-Interactive Journal. 9-CLASS-STANDARD
IX.1. In are distances from ( ) to sides of triangle. Prove that:
√ √ √ √
Proposed by Daniel Sitaru,Gabriel Tica -Romania
IX.2. In the following relationship holds:
(∑( )
)(∑
( )
) ∑
( )
Proposed by Daniel Sitaru,Cătălin Pană -Romania
IX.3. In the following relationship holds: √ ( )
( )( )
Proposed by Daniel Sitaru,Alina Tigae -Romania
IX.4. In acute – orthocenter, – incenter the following relationship holds:
4
5 4
5
( )
Proposed by Daniel Sitaru,Ramona Nălbaru -Romania
IX.5. If then:
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( )
( )( )
Proposed by Daniel Sitaru,Claudiu Ciulcu -Romania
IX.6. If ( ) then in any triangle having the area the following
inequality holds:
√( )( ) √( )( )
√( )( ) √
Proposed by D.M. Bătinețu – Giurgiu, Claudia Nănuți – Romania
IX.7. If ( ) and is a triangle having the area then:
( )
( )
( )
( ) √
Proposed by D.M. Bătinețu-Giurgiu – Romania, Martin Lukarevski - Macedonia
IX.8. In triangle having the area the following inequality holds:
( ) ( )
( ) √
Proposed by D.M. Bătinețu-Giurgiu, Nicolae Radu – Romania
IX.9. If , ) and is a triangle having the area , then:
.
/
.
/
.
/
(√ )
Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania
IX.10. If ( ) then in any triangle having the area the following
inequality holds:
4
54
( ) 5
Proposed by D.M. Bătinețu-Giurgiu,Nineta Oprescu– Romania
IX.11. If ( ) then in any triangle having the area the following
inequality holds:
4
5 4( ) ( )
5
Proposed by D.M. Bătinețu-Giurgiu,Dorina Goiceanu – Romania
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IX.12. In any triangle the following inequality holds:
where is the area of the triangle.
Proposed by D.M. Bătinețu – Giurgiu,Virginia Grigorescu – Romania
IX.13. If , ) and is a triangle having the area , then:
( )
( )
( ) √
Proposed by D.M. Bătinețu-Giurgiu, Dan Nănuți – Romania
IX.14. If , ) and then in triangle having the area the following