X2 t08 04 inequality techniques (2013)

68
Types of Proof

Transcript of X2 t08 04 inequality techniques (2013)

Page 1: X2 t08 04 inequality techniques (2013)

Types of Proof

Page 2: X2 t08 04 inequality techniques (2013)

Types of Proof 1. Deductive Proof

Start with known facts and deduce what you are trying to prove.

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Types of Proof 1. Deductive Proof

Start with known facts and deduce what you are trying to prove.

2. Inductive Proof

Assume what you are trying to prove and induce a solution.

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Types of Proof 1. Deductive Proof

Start with known facts and deduce what you are trying to prove.

2. Inductive Proof

Assume what you are trying to prove and induce a solution.

3. Proof by Contradiction

Assume the opposite of what you are trying to prove and create a

contradiction.

Contradiction means the original assumption is incorrect, therefore

the opposite must be true.

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Inequality Techniques

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Inequality Techniques 0 prove easier to becan it , prove To yxyx

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Inequality Techniques 0 prove easier to becan it , prove To yxyx

2

Prove 1995 e.g.22 qp

pqi

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Inequality Techniques 0 prove easier to becan it , prove To yxyx

2

Prove 1995 e.g.22 qp

pqi

pqqp

2

22

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Inequality Techniques 0 prove easier to becan it , prove To yxyx

2

Prove 1995 e.g.22 qp

pqi

pqqp

2

22

2

2 22 qpqp

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Inequality Techniques 0 prove easier to becan it , prove To yxyx

2

Prove 1995 e.g.22 qp

pqi

pqqp

2

22

2

2 22 qpqp

2

2qp

0

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Inequality Techniques 0 prove easier to becan it , prove To yxyx

2

Prove 1995 e.g.22 qp

pqi

pqqp

2

22

2

2 22 qpqp

2

2qp

0

pqqp

2

22

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Inequality Techniques 0 prove easier to becan it , prove To yxyx

2

Prove 1995 e.g.22 qp

pqi

pqqp

2

22

2

2 22 qpqp

2

2qp

0

pqqp

2

22

OR 2 2

Assume 2

p qpq

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Inequality Techniques 0 prove easier to becan it , prove To yxyx

2

Prove 1995 e.g.22 qp

pqi

pqqp

2

22

2

2 22 qpqp

2

2qp

0

pqqp

2

22

OR 2 2

Assume 2

p qpq

2 22pq p q

2 2

2

0 2

0 ( )

p pq q

p q

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Inequality Techniques 0 prove easier to becan it , prove To yxyx

2

Prove 1995 e.g.22 qp

pqi

pqqp

2

22

2

2 22 qpqp

2

2qp

0

pqqp

2

22

OR 2 2

Assume 2

p qpq

2 22pq p q

2 2

2

0 2

0 ( )

p pq q

p q

2But ( ) 0p q

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Inequality Techniques 0 prove easier to becan it , prove To yxyx

2

Prove 1995 e.g.22 qp

pqi

pqqp

2

22

2

2 22 qpqp

2

2qp

0

pqqp

2

22

OR 2 2

Assume 2

p qpq

2 22pq p q

2 2

2

0 2

0 ( )

p pq q

p q

2But ( ) 0p q 2 2

2

p qpq

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acbcabcbaii 222 Prove a) 1994

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acbcabcbaii 222 Prove a) 1994

02ba

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acbcabcbaii 222 Prove a) 1994

02ba

02 22 baba

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acbcabcbaii 222 Prove a) 1994

02ba

02 22 baba

abba 222

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acbcabcbaii 222 Prove a) 1994

02ba

02 22 baba

abba 222

bccb

acca

2

2

22

22

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acbcabcbaii 222 Prove a) 1994

02ba

02 22 baba

abba 222

bccb

acca

2

2

22

22

bcacabcba 222222 222

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acbcabcbaii 222 Prove a) 1994

02ba

02 22 baba

abba 222

bccb

acca

2

2

22

22

bcacabcba 222222 222

bcacabcba 222

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acbcabcbaii 222 Prove a) 1994

02ba

02 22 baba

abba 222

bccb

acca

2

2

22

22

bcacabcba 222222 222

bcacabcba 222

3

1 prove ,1 If b) bcacabcba

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acbcabcbaii 222 Prove a) 1994

02ba

02 22 baba

abba 222

bccb

acca

2

2

22

22

bcacabcba 222222 222

bcacabcba 222

3

1 prove ,1 If b) bcacabcba

bcacabcbacba 22222

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acbcabcbaii 222 Prove a) 1994

02ba

02 22 baba

abba 222

bccb

acca

2

2

22

22

bcacabcba 222222 222

bcacabcba 222

3

1 prove ,1 If b) bcacabcba

bcacabcbacba 22222

bcacabbcacabcba 22

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acbcabcbaii 222 Prove a) 1994

02ba

02 22 baba

abba 222

bccb

acca

2

2

22

22

bcacabcba 222222 222

bcacabcba 222

3

1 prove ,1 If b) bcacabcba

bcacabcbacba 22222

bcacabbcacabcba 22

23 cbabcacab

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acbcabcbaii 222 Prove a) 1994

02ba

02 22 baba

abba 222

bccb

acca

2

2

22

22

bcacabcba 222222 222

bcacabcba 222

3

1 prove ,1 If b) bcacabcba

bcacabcbacba 22222

bcacabbcacabcba 22

23 cbabcacab

3

1

13

bcacab

bcacab

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3

3

1 Prove c) abccba

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3

3

1 Prove c) abccba

bcacabcba 222

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3

3

1 Prove c) abccba

bcacabcba 222

0222 bcacabcba

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3

3

1 Prove c) abccba

bcacabcba 222

0222 bcacabcba

0222 bcacabcbacba

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3

3

1 Prove c) abccba

bcacabcba 222

0222 bcacabcba

0222 bcacabcbacba

022322

2223222223

bcacabcccbca

cbabcabbcbbaabccabaacaba

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3

3

1 Prove c) abccba

bcacabcba 222

0222 bcacabcba

0222 bcacabcbacba

022322

2223222223

bcacabcccbca

cbabcabbcbbaabccabaacaba

03333 abccba

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3

3

1 Prove c) abccba

bcacabcba 222

0222 bcacabcba

0222 bcacabcbacba

022322

2223222223

bcacabcccbca

cbabcabbcbbaabccabaacaba

03333 abccba

abccba 333

3

1

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3

3

1 Prove c) abccba

bcacabcba 222

0222 bcacabcba

0222 bcacabcbacba

022322

2223222223

bcacabcccbca

cbabcabbcbbaabccabaacaba

03333 abccba

abccba 333

3

1

3

1

3

1

3

1

,,let ccbbaa

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3

3

1 Prove c) abccba

bcacabcba 222

0222 bcacabcba

0222 bcacabcbacba

022322

2223222223

bcacabcccbca

cbabcabbcbbaabccabaacaba

03333 abccba

abccba 333

3

1

3

1

3

1

3

1

,,let ccbbaa

3

1

3

1

3

1

3

1cbacba

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3

3

1 Prove c) abccba

bcacabcba 222

0222 bcacabcba

0222 bcacabcbacba

022322

2223222223

bcacabcccbca

cbabcabbcbbaabccabaacaba

03333 abccba

abccba 333

3

1

3

1

3

1

3

1

,,let ccbbaa

3

1

3

1

3

1

3

1cbacba

3

3

1abccba

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nn

n aaan

aaa

21

21

Mean GeometricMean Arithmetic

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nn

n aaan

aaa

21

21

Mean GeometricMean Arithmetic

1 prove ,8111 Suppose d) xyzzyx

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nn

n aaan

aaa

21

21

Mean GeometricMean Arithmetic

1 prove ,8111 Suppose d) xyzzyx

81

8111

xyzyzxzzxyyx

zyx

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nn

n aaan

aaa

21

21

Mean GeometricMean Arithmetic

1 prove ,8111 Suppose d) xyzzyx

81

8111

xyzyzxzzxyyx

zyx

3

3

1xyzzyx AM GM

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nn

n aaan

aaa

21

21

Mean GeometricMean Arithmetic

1 prove ,8111 Suppose d) xyzzyx

81

8111

xyzyzxzzxyyx

zyx

3

3

1xyzzyx

33 xyzzyx

AM GM

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nn

n aaan

aaa

21

21

Mean GeometricMean Arithmetic

1 prove ,8111 Suppose d) xyzzyx

81

8111

xyzyzxzzxyyx

zyx

3

3

1xyzzyx

33 xyzzyx

33 xzyzxyxzyzxy

AM GM

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nn

n aaan

aaa

21

21

Mean GeometricMean Arithmetic

1 prove ,8111 Suppose d) xyzzyx

81

8111

xyzyzxzzxyyx

zyx

3

3

1xyzzyx

33 xyzzyx

33 xzyzxyxzyzxy

3 2223 zyxxzyzxy

AM GM

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nn

n aaan

aaa

21

21

Mean GeometricMean Arithmetic

1 prove ,8111 Suppose d) xyzzyx

81

8111

xyzyzxzzxyyx

zyx

3

3

1xyzzyx

33 xyzzyx

33 xzyzxyxzyzxy

3 2223 zyxxzyzxy

233 xyzxzyzxy

AM GM

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81 xyzyzxzxyzyx

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81 xyzyzxzxyzyx

83312

33 xyzxyzxyz

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81 xyzyzxzxyzyx

83312

33 xyzxyzxyz

83313

32

33 xyzxyzxyz

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81 xyzyzxzxyzyx

83312

33 xyzxyzxyz

83313

32

33 xyzxyzxyz

813

3 xyz

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81 xyzyzxzxyzyx

83312

33 xyzxyzxyz

83313

32

33 xyzxyzxyz

813

3 xyz

21 3 xyz

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81 xyzyzxzxyzyx

83312

33 xyzxyzxyz

83313

32

33 xyzxyzxyz

813

3 xyz

21 3 xyz

13 xyz

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81 xyzyzxzxyzyx

83312

33 xyzxyzxyz

83313

32

33 xyzxyzxyz

813

3 xyz

21 3 xyz

13 xyz

1xyz

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OR 1 2x x AM GM

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OR 1 2x x

1 2y y

1 2z z

AM GM

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OR 1 2x x

1 2y y

1 2z z

(1 )(1 ) 1 2 2 2

8

x y z x y z

xyz

AM GM

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OR 1 2x x

1 2y y

1 2z z

(1 )(1 ) 1 2 2 2

8

x y z x y z

xyz

8 8 xyz

AM GM

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OR 1 2x x

1 2y y

1 2z z

(1 )(1 ) 1 2 2 2

8

x y z x y z

xyz

8 8 xyz

1 1

1

1

xyz

xyz

xyz

AM GM

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9 2 2 2 1 1 1

Prove iiia b c a b b c a c a b c

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9 2 2 2 1 1 1

Prove iiia b c a b b c a c a b c

1 1 1

2 2

4

a b aba b ab

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9 2 2 2 1 1 1

Prove iiia b c a b b c a c a b c

1 1 1

2 2

4

a b aba b ab

1 1 4

a b a b

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9 2 2 2 1 1 1

Prove iiia b c a b b c a c a b c

1 1 1

2 2

4

a b aba b ab

1 1 4

a b a b

1 1 4

1 1 4

b c b c

a c a c

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9 2 2 2 1 1 1

Prove iiia b c a b b c a c a b c

1 1 1

2 2

4

a b aba b ab

1 1 4

a b a b

1 1 4

1 1 4

b c b c

a c a c

2 2 2 4 4 4

a b c a b b c a c

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9 2 2 2 1 1 1

Prove iiia b c a b b c a c a b c

1 1 1

2 2

4

a b aba b ab

1 1 4

a b a b

1 1 4

1 1 4

b c b c

a c a c

2 2 2 4 4 4

a b c a b b c a c

1 1 1 2 2 2

a b c a b b c a c

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9 2 2 2 1 1 1

Prove iiia b c a b b c a c a b c

1 1 1

2 2

4

a b aba b ab

1 1 4

a b a b

1 1 4

1 1 4

b c b c

a c a c

2 2 2 4 4 4

a b c a b b c a c

1 1 1 2 2 2

a b c a b b c a c

3 31 1 1 1

3 3

9

a b c abca b c abc

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9 2 2 2 1 1 1

Prove iiia b c a b b c a c a b c

1 1 1

2 2

4

a b aba b ab

1 1 4

a b a b

1 1 4

1 1 4

b c b c

a c a c

2 2 2 4 4 4

a b c a b b c a c

1 1 1 2 2 2

a b c a b b c a c

3 31 1 1 1

3 3

9

a b c abca b c abc

1 1 1 9

a b c a b c

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9 2 2 2 1 1 1

Prove iiia b c a b b c a c a b c

1 1 1

2 2

4

a b aba b ab

1 1 4

a b a b

1 1 4

1 1 4

b c b c

a c a c

2 2 2 4 4 4

a b c a b b c a c

1 1 1 2 2 2

a b c a b b c a c

3 31 1 1 1

3 3

9

a b c abca b c abc

1 1 1 9

a b c a b c

1 1 1 9

2

2 2 2 9

a b b c a c a b c

a b b c a c a b c

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9 2 2 2 1 1 1

Prove iiia b c a b b c a c a b c

1 1 1

2 2

4

a b aba b ab

1 1 4

a b a b

1 1 4

1 1 4

b c b c

a c a c

2 2 2 4 4 4

a b c a b b c a c

1 1 1 2 2 2

a b c a b b c a c

3 31 1 1 1

3 3

9

a b c abca b c abc

1 1 1 9

a b c a b c

1 1 1 9

2

2 2 2 9

a b b c a c a b c

a b b c a c a b c

9 2 2 2 1 1 1

a b c a b b c a c a b c

Page 68: X2 t08 04 inequality techniques (2013)

9 2 2 2 1 1 1

Prove iiia b c a b b c a c a b c

1 1 1

2 2

4

a b aba b ab

1 1 4

a b a b

1 1 4

1 1 4

b c b c

a c a c

2 2 2 4 4 4

a b c a b b c a c

1 1 1 2 2 2

a b c a b b c a c

3 31 1 1 1

3 3

9

a b c abca b c abc

1 1 1 9

a b c a b c

1 1 1 9

2

2 2 2 9

a b b c a c a b c

a b b c a c a b c

9 2 2 2 1 1 1

a b c a b b c a c a b c

Inequalities Sheet

Exercise 10D

Note: Cambridge 8H (Book 1); 28