SOLUTION OF ONE CONJECTURE ON
INEQUALITIES WITH POWER-EXPONENTIAL
FUNCTIONS
Solution Of One Conjecture
Ladislav Matejíčka
vol. 10, iss. 3, art. 72, 2009
LADISLAV MATEJÍČKA
Institute of Information Engineering, Automation and Mathematics
Faculty of Chemical Food Technology
Slovak University of Technology in Bratislava
Slovakia
EMail: matejicka@tnuni.sk
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Received:
20 July, 2009
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Accepted:
24 August, 2009
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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. Cîrtoaje, 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 Cîrtoaje for his valuable remarks, suggestions and for his improving some inequalities in the paper.
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Contents
1
Introduction
3
2
Proof of Conjecture 4.6
4
Solution Of One Conjecture
Ladislav Matejíčka
vol. 10, iss. 3, art. 72, 2009
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1.
Introduction
In the paper [1], V. Cîrtoaje posted 5 conjectures on inequalities with power-exponential
functions. In this paper, we prove Conjecture 4.6.
Conjecture 4.6. Let r 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.
Solution Of One Conjecture
Ladislav Matejíčka
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 − x1 , b = x1 , r = 3x for x > 1.
Then we have
arb + bra > 2.
(2.1)
In fact,
Solution Of One Conjecture
3x(2− x1 )
6x−3
1
1
12
6
1
+
=8−
+ 2− 3+
x
x
x
x
x
6x−3
and if we show that x1
> −6 + 12
− x62 + x13 then the inequality (2.1) will be
x
fulfilled for all x > 1. Put t = x1 , then 0 < t < 1. The inequality (2.1) becomes
1
2−
x
3
6
t
Ladislav Matejíčka
vol. 10, iss. 3, art. 72, 2009
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t > t3 (t3 − 6t2 + 12t − 6) = t3 β(t),
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where β(t) = t3 − 6t2 + 12t − 6. From β 0 (t) = 3(t − 2)2 , β(0) = −6, and from that
there is only one real t0 = 0.7401 such that β(t0 ) = 0 and we have that β(t) ≤ 0 for
6
0 ≤ t ≤ t0 . Thus, it suffices to show that t t > t3 β(t) for t0 < t < 1. Rewriting the
previous inequality we get
6
α(t) =
− 3 ln t − ln(t3 − 6t2 + 12t − 6) > 0.
t
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From α(1) = 0, it suffices to show that α0 (t) < 0 for t0 < t < 1, where
6
6
1
3t2 − 12t + 12
0
α (t) = − 2 ln t +
−3
− 3
.
t
t
t t − 6t2 + 12t − 6
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α0 (t) < 0 is equivalent to
t2 (t − 2)2
> 0.
t3 − 6t2 + 12t − 6
From γ(1) = 0, it suffices to show that γ 0 (t) < 0 for t0 < t < 1, where
γ(t) = 2 ln t − 2 + t +
(4t3 − 12t2 + 8t)(t3 − 6t2 + 12t − 6) − (t4 − 4t3 + 4t2 )(3t2 − 12t + 12)
(t3 − 6t2 + 12t − 6)2
2
+ +1
t
t6 − 12t5 + 56t4 − 120t3 + 120t2 − 48t 2
=
+ + 1.
(t3 − 6t2 + 12t − 6)2
t
γ 0 (t) =
Solution Of One Conjecture
Ladislav Matejíčka
vol. 10, iss. 3, art. 72, 2009
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0
γ (t) < 0 is equivalent to
7
6
Contents
5
4
3
2
p(t) = 2t − 22t + 92t − 156t + 24t + 240t − 252t + 72 < 0.
From
p(t) = 2(t − 1)(t6 − 10t5 + 36t4 − 42t3 − 30t2 + 90t − 36),
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it suffices to show that
q(t) = t6 − 10t5 + 36t4 − 42t3 − 30t2 + 90t − 36 > 0.
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Since q(0.74) = 5.893, q(1) = 9 it suffices to show that q 00 (t) < 0 and (2.2) will be
proved. Indeed, for t0 < t < 1, we have
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(2.2)
00
4
3
2
q (t) = 2(15t − 100t + 216t − 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 = 1 − x 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 show
that
(2.3)
(1 − x)r(1+x) + (1 + x)r(1−x) ≤ 2 for
0 < x < 1, 0 < r ≤ 3.
Denote ϕ(x) = (1 − x)r(1+x) + (1 + x)r(1−x) . We show that ϕ0 (x) < 0 for 0 < x < 1,
0 < r ≤ 3 which gives that (2.3) is valid (ϕ(0) = 2).
1+x
1−x
0
r(1+x)
r(1−x)
ϕ (x) = (1−x)
r ln(1 − x) − r
+(1+x)
r
− r ln(1 + x) .
1−x
1+x
Solution Of One Conjecture
Ladislav Matejíčka
vol. 10, iss. 3, art. 72, 2009
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0
The inequality ϕ (x) < 0 is equivalent to
r 1−x
1+x
1+x
2 rx
(2.4)
− ln(1 + x) ≤ (1 − x )
− ln(1 − x) .
1−x
1+x
1−x
1−x
−ln(1+x)
1+x
0
2
1
− (1+x)
2 − 1+x
If δ(x) =
≤ 0, then (2.4) is evident. Since δ (x) =
<0
for 0 ≤ x < 1, δ(0) = 1 and δ(1) = − ln 2, we have δ(x) > 0 for 0 ≤ x < x0 ∼
=
0.4547. Therefore, it suffices to show that h(x) ≥ 0 for 0 ≤ x ≤ x0 , where
1+x
1+x
2
h(x) = rx ln(1 − x ) − r ln
+ ln
− ln(1 − x)
1−x
1−x
1−x
− ln
− ln(1 + x) .
1+x
We show that h0 (x) ≥ 0 for 0 < x < x0 , 0 < r ≤ 3. Then from h(0) = 0 we obtain
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h(x) ≥ 0 for 0 < x ≤ x0 and it implies that the inequality (2.4) is valid.
h0 (x) = r ln(1 − x2 ) − 2r
1 + x2
3−x
+
2
1−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 h0 (x) ≥ 0, 0 < x < x0 is
equivalent to
(2.5) r(2x2 + 2 − (1 − x2 )(A + B)) ≤
3 + 2x − x2
3 − 2x − x2
+
.
1 − x − (1 + x)A 1 + x − (1 − x)B
Solution Of One Conjecture
Ladislav Matejíčka
vol. 10, iss. 3, art. 72, 2009
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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
3 + 2x − x2
+
1 − x − (1 + x)A 1 + x − (1 − x)B
and then the inequality (2.5) will be fulfilled for 0 < r ≤ 3. The inequality (2.6) for
0 < x < x0 is equivalent to
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2
4
4
3
2
4
3
2
(2.7) 6x −6x −(9x +13x +5x +7x+6)A−(9x −13x +5x −7x+6)B
− (3x4 + 6x3 − 6x − 3)A2 − (3x4 − 6x3 + 6x − 3)B 2
− (12x4 − 12)AB − (3x4 − 6x2 + 3)AB(A + B) ≤ 0.
It is easy to show that the following Taylor’s formulas are valid for 0 < x < 1:
A=
∞
X
(−1)n
n=0
n+1
x
n+1
,
B=−
∞
X
n=0
1
xn+1 ,
n+1
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A2 =
∞
X
2(−1)n+1
n=1
n
X
1
∞
X
!
n
X1
2
xn+1 , B 2 =
n+1
i
n + 1 i=1 i
n=1
i=1
!
2n+1
∞
X (−1)i+1
X
1
AB = −
x2n+2 .
n + 1 i=1
i
n=0
Since
4
A2 + B 2 =
n+1
n=1,3,5,...
X
and
4
n+1
n
X
1
i=1
!
i
4
≤
n+1
n
X
1
i=1
!
xn+1 ,
!
i
n−1
1+
2
Solution Of One Conjecture
xn+1
Ladislav Matejíčka
vol. 10, iss. 3, art. 72, 2009
= 2,
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we have
A2 + B 2 =
4
n+1
n=1,3,5,...
X
n
X
1
i=1
i
!
xn+1
n
X
X
11 4 137 6
4
1
= 2x + x +
x +
6
90
n + 1 i=1 i
n=7,9,...
X
11
137 6
< 2x2 + x4 +
x +2
xn+1
6
90
n=7,9,...
2
11
137 6
2x8
= 2x + x4 +
x +
.
6
90
1 − x2
From this and from the previous Taylor’s formulas we have
1 4 1 6 1
x8
2
(2.8)
A + B > −x − x − x −
,
2
3
4 1 − x2
2
!
xn+1
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(2.9)
(2.10)
2
2
2
A − B > 2x + x3 + x5 + x7 ,
3
5
7
A2 + B 2 < 2x2 +
2x8
11 4 137 6
x +
x +
,
6
90
1 − x2
Solution Of One Conjecture
Ladislav Matejíčka
(2.11)
5
A2 − B 2 < −2x3 − x5 ,
3
vol. 10, iss. 3, art. 72, 2009
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(2.12)
AB < −x2 −
5 4
x
12
for
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0 < x < 1.
Now, having in view (2.12) and the obvious inequality A + B < 0, to prove (2.7) it
suffices to show that
6x2 − 6x4 − (6 + 5x2 + 9x4 )(A + B) + (7x + 13x3 )(B − A) + (3 − 3x4 )(A2 + B 2 )
5 4
3
2
2
4
2
+ (6x − 6x )(A − B ) − (12 − 12x ) x + x
12
5
+ x2 + x4 (3 − 6x2 + 3x4 )(A + B) ≤ 0.
12
By using the inequalities (2.10), (2.11), the previous inequality will be proved if we
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show that
6x2 − 6x4 − (6 + 5x2 + 9x4 )(A + B) + (7x + 13x3 )(B − A)
11 4 137 6
2x8
5 5
4
2
3
3
+ (3 − 3x ) 2x + x +
x +
− (6x − 6x ) 2x + x
6
90
1 − x2
3
5
5
− (12 − 12x4 ) x2 + x4 + x2 + x4 (3 − 6x2 + 3x4 )(B + A) ≤ 0,
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12
Ladislav Matejíčka
which can be rewritten as
(2.13)
−
vol. 10, iss. 3, art. 72, 2009
35 4 377 6 19 8 137 10
x +
x + x −
x + 6(x8 + x10 )
2
30
2
30
55 4 1 6 5 8
2
− (A + B) 6 + 2x + x − x − x
4
2
4
+ (7x + 13x3 )(B − A) ≤ 0.
To prove (2.13) it suffices to show
(2.14)
− 8x2 −
259 4 357 6 1841 8 337 10 19 12
5
x +
x +
x +
x − x − x14
6
20
120
420
24
12
x8
3 1 2 55 4 1 6
5 8
+
+ x + x − x − x < 0.
1 − x2 2 2
16
8
16
It follows from (2.8) and (2.9). Since 0 < x <
ε(x) = −8x2 −
Solution Of One Conjecture
1
2
we have
1
1−x2
< 43 . If we show
259 4 357 6 1841 8 337 10 19 12
x +
x +
x +
x − x
6
20
120
420
24
5 14
2
55
1
5
− x + 2x8 + x10 + x12 − x14 − x16 < 0,
12
3
12
6
12
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then the inequality (2.14) will be proved. From x6 < x4 , x8 < x4 , x10 < x4 and
x12 < x4 , we obtain that
ε(x) < −8x2 −
19 4
7
5
x − x14 − x16 < 0.
7
12
12
This completes the proof.
Solution Of One Conjecture
Ladislav Matejíčka
vol. 10, iss. 3, art. 72, 2009
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References
[1] V. CÎRTOAJE, On some inequalities with power-exponential functions, J. Inequal. Pure Appl. Math., 10(1) (2009), Art. 21. [ONLINE: http://jipam.
vu.edu.au/article.php?sid=1077]
Solution Of One Conjecture
Ladislav Matejíčka
vol. 10, iss. 3, art. 72, 2009
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