8/3/2019 OPEN INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS Conjecture 4.7-4.8

1/4

Solutions of Two Conjectures on Inequalities with

Power-exponential Functions

Yin Li

Abstract: In this paper, we mainly prove two conjectures posted by V.Crtoaje. Related problems are also presented.

Key Words: Bernoullis inequality; Monotonicity; Concave function; Power-exponential Functions

2000 AMS. Sub. Class: 26D10

Email: yinli79 a163.com

1. Introduction

In 2006, A. Zeikii posted and proved on the Mathlinks Forum the followinginequality

aa + bb ab + ba (1.1)where a and b are positive real numbers less than or equal to 1. In addition, heconjectured that the following inequality holds under the same conditions

a2a + b2b a2b + b2a. (1.2)Later, V. Crtoaje proposed and proved the open inequality

aea + beb aeb + bea (1.3)for either a b 1

eor 1

e a b > 0 in [1]. In the same paper, he also posted

several conjectures.Recently, L. Matejcka proved Conjecture 4.6 posted by V. Crtoa je in [2].In this paper, we mainly prove Conjecture 4.7 and Conjecture 4.8. In addi-

tion, we give a partial solution to Conjecture 4.3. Finally, other related openproblems are presented.

2. Main Results

Conjecture 4.7. If a and b are nonnegative real numbers such that a + b = 2,then

a3b + b3a + (a b

2)4 2. (2.1)

Department of Mathematics and Information Science, Binzhou University,Shandong

256600

1

8/3/2019 OPEN INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS Conjecture 4.7-4.8

2/4

Proof. Without loss of generality, assume that a b.Case 0.3 b 1. Since 0.7 a 1 1, by Bernoullis inequality we have

ab 1 + b(a 1) = 1 + b + b2

andba = bba1 b[1 + (a 1)(b 1)] = b2(2 b).

Therefore

a3b + b3a + (a b

2)4 2

(1 + b + b2)3 + b6(2 b)3 + (1 b)4 2 b(b 1)3(b5 3b4 + 3b2 + 3b 1) 0

.

(In fact, we denote g(b) = b5

3b4 + 3b2 + 3b

1. Since g(0) =

1 < 0 and

g(0.3) = 0.14813 > 0, then g(b) = 0 only exist a real root (0, 0.3) by theoremof zero point and simple argument.)Case 0 b < 19 . Since

a3b (1 + b)3 = 1 + 3b + 3b2 + b3

and b3a 8b6, we have

a3b + b3a + (a b

2)4 2

b + 9b2 3b3 + b4 + 8b6 0.

Case 1

9 b < 0.3. Defining function

f(b) = (2 b)3b + b3(2b) + (1 b)4 2,

we havef(b) b(b 1)3g(b)

where g(b) = b5 3b4 + 3b2 + 3b 1. [See Case 0.3 b 1.]Owing to g() = 0, equality f(b) = 0 dont have real root on [ 1

9, ] [1

9, 0.3].

Hence f(b) f( 19

) < 0 or f(b) f() f(0.3) < 0. This complete the proof.Conjecture 4.8. If a and b are nonnegative real numbers such that a + b = 1,then

a2b + b2a 1. (2.2)

Proof. Without loss of generality, assume that a b.Case 0 a 1

2. In order to prove above inequality, we show that f(a) 1

2,

where f(a) = a2b = a2(1a). Since

f(a) = a2(1a)g(a) > 0,

where g(a) = 2a 2 2lna. (In fact, since g(a) = 2

a2 2

a< 0, g(a) is strictly

decreasing. Hence, we get g(a) g(1) = 0 and f(a) > 0.)

Therefore, f(a) is strictly increasing as a [0, 12

], and then f(a) = a2b f( 12) =

12

. On the other hand, proof of b2a 12 is similar to the previous case.

2

8/3/2019 OPEN INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS Conjecture 4.7-4.8

3/4

So, we have a2b + b2a 12

+ 12

= 1.

Case12 a 1. Since f(a) = a

2(1a)

, we have

f(a) = a2(1a)2

a 2 2lna

and

f(a) = a2(1a)[(2(1 a)

a 2 ln a)2 2

a2 2

a]

by a simple calculation.

Next, we show that f(a) 0. It suffices to show

(2(1 a)

a 2 ln a)2 2

a2+

2

a

or

1 a a ln a

2 + 2a

2.

Owing to h(a) = 2 lna, where h(a) = 1aalna, h(a) is strictly decreasingas a [ 1

e2, 1], and then h(a) h( 1

2) = 1

2+ 1

2ln 2.

On the other hand,2+2a2

32

> 12

+ 12

ln2 is easy to prove in a [12

, 1]. Therefore, we easily know that f(a) is a concave function. Using Jessensinequality, we have

f(a) + f(b) = a2b + b2a 2f(a + b

2) = 2f(

1

2) = 1.

Consider case 0 a 12

and case 12 a 1, we complete the proof.

In order to give a partial proof to Conjecture 4.3, we give following lemma.Lemma 2.1[1]. If 0 < r 2, then

ara + brb arb + bra (2.3)

holds for all positive real numbers a and b.Proposition 2.1. If a,b,c satisfy max {a , b, c} = a 1 or 12 c b a < 1,then

a2a + b2b + c2c a2b + b2c + c2a. (2.4)

Proof. Case max {a,b,c} = a 1. By Lemma 2.1, we have

b2b + c2c b2c + c2b.

Thus it suffice to provea2a + c2b a2b + c2a.

For a = b, this inequality is an equality. Otherwise, for a > b, we substitutex = c2b, y = a2b and s = a

b(where x y, y 1, and s 1) to rewrite the

inequality as f(x) 0 where

f(y) = ys + x y xs.

3

8/3/2019 OPEN INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS Conjecture 4.7-4.8

4/4

Since

f(y) = sys

1

1 s 1 0,then f(y) is strictly increasing for x y, and therefore f(y) f(x) = 0.Case 1

2 c b a < 1. By Lemma 2.1, we have

a2a + b2b a2b + b2a.

Thus it suffice to proveb2a + c2c b2c + c2a,

which is equivalent to g(b) g(c), where g(x) = x2a x2c. The inequality istrue if g(x) 0 for c x b. From

g(x) = 2ax2a1 2cx2c1

= 2c x2c1c2a2c(a c12a+2c),

we need to show that ac12a+2c 0. Since 0 1 2a + 2c 1, by Bernoullisinequality, we have

a c12a+2c a (c 1)(1 + (1 2a + 2c))= (a c)(2c 1) 0 .

We finish the proof of Proposition 2.1.

In this way, we give a partial solution of Conjecture 4.3 in [1].

3. Open Problems

Problem 3.1. If a,b,c are positive real numbers, then

aaa

+ bbb

+ ccc

abc

+ bca

+ cab

. (3.1)

Problem 3.2. If a,b,c are positive real numbers, then

aaa

bbb

ccc

abc

bca

cab

. (3.2)

References

[1] V.Crtoaje, On some inequalities with power-exponential functions, J. In-equal. Pure Appl. Math.10(1)(2009), Art. 21.

[2] L.Matejcka, Solution of one conjecture on inequalities with power-exponential functions, J. Inequal. Pure Appl. Math.10(3)(2009), Art. 72.

4