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8/3/2019 OPEN INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS Conjecture4.6
1/12
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.
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Solution Of One Conjecture
Ladislav Matejcka
vol. 10, iss. 3, art. 72, 2009
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Contents
1 Introduction 3
2 Proof of Conjecture 4.6 4
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Solution Of One Conjecture
Ladislav Matejcka
vol. 10, iss. 3, art. 72, 2009
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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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Solution Of One Conjecture
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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Solution Of One Conjecture
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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Solution Of One Conjecture
Ladislav Matejcka
vol. 10, iss. 3, art. 72, 2009
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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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Solution Of One Conjecture
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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Solution Of One Conjecture
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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Solution Of One Conjecture
Ladislav Matejcka
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