open inequalities with power-exponential functions conjecture4.6

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Solution Of One Conjecture

Ladislav Matejcka

vol. 10, iss. 3, art. 72, 2009

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SOLUTION OF ONE CONJECTURE ONINEQUALITIES WITH POWER-EXPONENTIAL

FUNCTIONS

LADISLAV MATEJCKAInstitute of Information Engineering, Automation and MathematicsFaculty of Chemical Food TechnologySlovak University of Technology in BratislavaSlovakia

EMail: [email protected]

Received: 20 July, 2009

Accepted: 24 August, 2009

Communicated by: S.S. Dragomir

2000 AMS Sub. Class.: 26D10.

Key words: Inequality, Power-exponential functions.Abstract: In this paper, we prove one conjecture presented in the paper [V. Crtoaje, On

some inequalities with power-exponential functions, J. Inequal. Pure Appl. Math.10 (2009) no. 1, Art. 21. http://jipam.vu.edu.au/article.php?sid=1077].

Acknowledgements: The author is deeply grateful to Professor Vasile Crtoaje for his valuable re-marks, suggestions and for his improving some inequalities in the paper.
http://jipam.vu.edu.au/article.php?sid=1077http://jipam.vu.edu.au/article.php?sid=1077mailto:[email protected]:[email protected]://jipam.vu.edu.au/http://close/http://fullscreen/http://goback/http://prevpage/http://lastpage/mailto:[email protected]://jipam.vu.edu.au/


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Ladislav Matejcka

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Contents

1 Introduction 3

2 Proof of Conjecture 4.6 4
http://jipam.vu.edu.au/http://close/http://fullscreen/http://goback/http://prevpage/http://lastpage/mailto:[email protected]://jipam.vu.edu.au/


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Ladislav Matejcka

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1. Introduction

Inthepaper[1], V. Crtoaje posted 5 conjectures on inequalities with power-exponentialfunctions. In this paper, we prove Conjecture 4.6.

Conjecture 4.6. Letr be a positive real number. The inequality

(1.1) arb + bra 2

holds for all nonnegative real numbers a and b with a + b = 2, if and only if r 3.


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Ladislav Matejcka

vol. 10, iss. 3, art. 72, 2009

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2. Proof of Conjecture 4.6

First, we prove the necessary condition. Put a = 2 1x

, b = 1x

, r = 3x for x > 1.Then we have

(2.1) arb + bra > 2.

In fact, 2

1

x

3+

1

x

3x(2 1x

)= 8

12

x+

6

x2

1

x3+

1

x

6x3

and if we show that

1x

6x3

> 6 + 12x

6x2

+ 1x3

then the inequality (2.1) will befulfilled for all x > 1. Put t = 1

x, then 0 < t < 1. The inequality (2.1) becomes

t6

t > t3(t3 6t2 + 12t 6) = t3(t),

where (t) = t3 6t2 + 12t 6. From (t) = 3(t 2)2, (0) = 6, and from thatthere is only one real t0 = 0.7401 such that (t0) = 0 and we have that (t) 0 for0 t t0. Thus, it suffices to show that t

6

t > t3(t) for t0 < t < 1. Rewriting theprevious inequality we get

(t) =

6t 3

ln t ln(t3 6t2 + 12t 6) > 0.

From (1) = 0, it suffices to show that (t) < 0 for t0 < t < 1, where

(t) = 6

t2ln t +

6

t 3

1

t

3t2 12t + 12

t3 6t2 + 12t 6.


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Ladislav Matejcka

vol. 10, iss. 3, art. 72, 2009

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(t) < 0 is equivalent to

(t) = 2 ln t 2 + t +t2(t 2)2

t3 6t2 + 12t 6> 0.

From (1) = 0, it suffices to show that (t) < 0 for t0 < t < 1, where

(t) = (4t3

12t2

+ 8t)(t3

6t2

+ 12t 6) (t4

4t3

+ 4t2

)(3t2

12t + 12)(t3 6t2 + 12t 6)2

+2

t+ 1

=t6 12t5 + 56t4 120t3 + 120t2 48t

(t3 6t2 + 12t 6)2+

2

t+ 1.

(t) < 0 is equivalent to

p(t) = 2t7 22t6 + 92t5 156t4 + 24t3 + 240t2 252t + 72 < 0.

Fromp(t) = 2(t 1)(t6 10t5 + 36t4 42t3 30t2 + 90t 36),

it suffices to show that

(2.2) q(t) = t6 10t5 + 36t4 42t3 30t2 + 90t 36 > 0.

Since q(0.74) = 5.893, q(1) = 9 it suffices to show that q(t) < 0 and (2.2) will beproved. Indeed, for t0 < t < 1, we have

q(t) = 2(15t4 100t3 + 216t2 126t 30)

< 2(40t4 100t3 + 216t2 126t 30)

= 4(t 1)(20t3 30t2 + 78t + 15)

< 4(t 1)(30t2 + 78t) < 0.


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This completes the proof of the necessary condition.We prove the sufficient condition. Put a = 1x and b = 1+x, where 0 < x < 1.

Since the desired inequality is true for x = 0 and for x = 1, we only need to showthat

(2.3) (1 x)r(1+x) + (1 + x)r(1x) 2 for 0 < x < 1, 0 < r 3.

Denote (x) = (1x)r(1+x) +(1+ x)r(1x). We show that (x) < 0 for 0 < x < 1,0 < r 3 which gives that (2.3) is valid ((0) = 2).

(x) = (1x)r(1+x)

r ln(1 x) r1 + x

1 x

+(1+x)r(1x)

r

1 x

1 + x r ln(1 + x)

.

The inequality (x) < 0 is equivalent to

(2.4)

1 + x

1 x

r

1 x

1 + x ln(1 + x)

(1 x2)rx

1 + x

1 x ln(1 x)

.

If(x) = 1x1+xln(1+x) 0, then (2.4) is evident. Since (x) = 2

(1+x)2

11+x

< 0

for 0 x < 1, (0) = 1 and (1) = ln 2, we have (x) > 0 for 0 x < x0 =0.4547. Therefore, it suffices to show that h(x) 0 for 0 x x0, where

h(x) = rx ln(1 x2) r ln

1 + x1 x

+ ln

1 + x1 x

ln(1 x)

ln

1 x

1 + x ln(1 + x)

.

We show that h(x) 0 for 0 < x < x0, 0 < r 3. Then from h(0) = 0 we obtain


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Ladislav Matejcka

vol. 10, iss. 3, art. 72, 2009

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h(x) 0 for 0 < x x0 and it implies that the inequality (2.4) is valid.

h(x) = r ln(1 x2) 2r1 + x2

1 x2+

3 x

(1 x)(1 + x (1 x)ln(1 x))

+3 + x

(1 + x)(1

x

(1 + x) ln(1 + x))

.

Put A = ln(1 + x) and B = ln(1 x). The inequality h(x) 0, 0 < x < x0 isequivalent to

(2.5) r(2x2 + 2 (1 x2)(A + B)) 3 2x x2

1 x (1 + x)A+

3 + 2x x2

1 + x (1 x)B.

Since 2x2 + 2 (1 x2)(A + B) > 0 for 0 < x < 1, it suffices to prove that

(2.6) 3(2x2 + 2 (1 x2)(A + B)) 3 2x x2

1 x (1 + x)A+

3 + 2x x2

1 + x (1 x)B

and then the inequality (2.5) will be fulfilled for 0 < r 3. The inequality (2.6) for0 < x < x0 is equivalent to

(2.7) 6x26x4(9x4+13x3+5x2+7x+6)A(9x413x3+5x27x+6)B

(3x4 + 6x3 6x 3)A2 (3x4 6x3 + 6x 3)B2

(12x4 12)AB (3x4 6x2 + 3)AB(A + B) 0.

It is easy to show that the following Taylors formulas are valid for 0 < x < 1:

A =n=0

(1)n

n + 1xn+1, B =

n=0

1

n + 1xn+1,


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Ladislav Matejcka

vol. 10, iss. 3, art. 72, 2009

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A2 =

n=1

2(1)n+1

n + 1

n

i=1

1

i

xn+1, B2 =

n=1

2

n + 1

n

i=1

1

i

xn+1,

AB =

n=0

1

n + 1

2n+1i=1

(1)i+1

i

x2n+2.

SinceA2 + B2 =

n=1,3,5,...

4

n + 1

n

i=1

1

i

xn+1

and4

n + 1

n

i=1

1

i

4

n + 1

1 +

n 1

2

= 2,

we have

A2 + B2 =

n=1,3,5,...

4

n + 1

n

i=1

1

i

xn+1

= 2x2 +11

6x4 +

137

90x6 +

n=7,9,...

4

n + 1

n

i=1

1

i

xn+1

< 2x2 +11

6

x4 +137

90

x6 + 2 n=7,9,...

xn+1

= 2x2 +11

6x4 +

137

90x6 +

2x8

1 x2.

From this and from the previous Taylors formulas we have

(2.8) A + B > x2 1

2x4

1

3x6

1

4

x8

1 x2

,


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vol. 10, iss. 3, art. 72, 2009

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(2.9) AB > 2x +2

3x3 +

2

5x5 +

2

7x7,

(2.10) A2 + B2 < 2x2 +11

6 x4 +137

90 x6 +2x8

1 x2 ,

(2.11) A2 B2 < 2x3 5

3x5,

(2.12) AB < x2 5

12x4 for

0< x

open inequalities with power-exponential functions conjecture4.6

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