ON SOME INEQUALITIES WITH
POWER-EXPONENTIAL FUNCTIONS
Inequalities With PowerExponential Functions
VASILE CÎRTOAJE
Department of Automation and Computers
University Of Ploieşti
Ploiesti, Romania
EMail: vcirtoaje@upg-ploiesti.ro
Vasile Cîrtoaje
vol. 10, iss. 1, art. 21, 2009
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Received:
13 October, 2008
Accepted:
09 January, 2009
Communicated by:
JJ
II
F. Qi
J
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2000 AMS Sub. Class.:
26D10.
Page 1 of 14
Key words:
Power-exponential function, Convex function, Bernoulli’s inequality, Conjecture.
Abstract:
In this paper, we prove the open inequality aea + beb ≥ aeb + bea for either
a ≥ b ≥ 1e or 1e ≥ a ≥ b > 0. In addition, other related results and conjectures
are presented.
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Contents
1
Introduction
3
2
Main Results
4
3
Proofs of Theorems
5
4
Other Related Inequalities
10
Inequalities With PowerExponential Functions
Vasile Cîrtoaje
vol. 10, iss. 1, art. 21, 2009
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1.
Introduction
In 2006, A. Zeikii posted and proved on the Mathlinks Forum [1] the following inequality
(1.1)
aa + b b ≥ ab + b a ,
where a and b are positive real numbers less than or equal to 1. In addition, he
conjectured that the following inequality holds under the same conditions:
Inequalities With PowerExponential Functions
a2a + b2b ≥ a2b + b2a .
vol. 10, iss. 1, art. 21, 2009
(1.2)
Starting from this, we have conjectured in [1] that
(1.3)
ea
eb
eb
a +b ≥a +b
for all positive real numbers a and b.
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ea
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2.
Main Results
In what follows, we will prove some relevant results concerning the power-exponential
inequality
(2.1)
ara + brb ≥ arb + bra
Inequalities With PowerExponential Functions
for a, b and r positive real numbers. We will prove the following theorems.
Theorem 2.1. Let r, a and b be positive real numbers. If (2.1) holds for r = r0 , then
it holds for any 0 < r ≤ r0 .
Vasile Cîrtoaje
vol. 10, iss. 1, art. 21, 2009
Theorem 2.2. If a and b are positive real numbers such that max{a, b} ≥ 1, then
(2.1) holds for any positive real number r.
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Theorem 2.3. If 0 < r ≤ 2, then (2.1) holds for all positive real numbers a and b.
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Theorem 2.4. If a and b are positive real numbers such that either a ≥ b ≥
1
≥ a ≥ b, then (2.1) holds for any positive real number r ≤ e.
r
1
r
or
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Theorem 2.5. If r > e, then (2.1) does not hold for all positive real numbers a and b.
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From the theorems above, it follows that the inequality (2.1) continues to be an
open problem only for 2 < r ≤ e and 0 < b < 1r < a < 1. For the most interesting
value of r, that is r = e, only the case 0 < b < 1e < a < 1 is not yet proved.
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3.
Proofs of Theorems
Proof of Theorem 2.1. Without loss of generality, assume that a ≥ b. Let x = ra
and y = rb, where x ≥ y. The inequality (2.1) becomes
xx − y x ≥ rx−y (xy − y y ).
(3.1)
Inequalities With PowerExponential Functions
By hypothesis,
xx − y x ≥ r0x−y (xy − y y ).
y
r0x−y (xy
y
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y
Since x − y ≥ 0 and x − y ≥ 0, we have
−y ) ≥ r
hence
xx − y x ≥ r0x−y (xy − y y ) ≥ rx−y (xy − y y ).
x−y
y
y
(x − y ), and
vol. 10, iss. 1, art. 21, 2009
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Proof of Theorem 2.2. Without loss of generality, assume that a ≥ b and a ≥ 1.
rb ra
From ar(a−b) ≥ br(a−b) , we get brb ≥ a arab . Therefore,
ra
rb
rb
a +b −a −b
ra
arb bra
≥ a + ra − arb − bra
a
(ara − arb )(ara − bra )
=
≥ 0,
ara
ra
because ara ≥ arb and ara ≥ bra .
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Proof of Theorem 2.3. By Theorem 2.1 and Theorem 2.2, it suffices to prove (2.1)
for r = 2 and 1 > a > b > 0. Setting c = a2b , d = b2b and s = ab (where c > d > 0
and s > 1), the desired inequality becomes
cs − ds ≥ c − d.
In order to prove this inequality, we show that
(3.2)
cs − ds > s(cd)
s−1
2
(c − d) > c − d.
The left side of the inequality in (3.2) is equivalent to f (c) > 0, where f (c) =
s−1
s−3
cs − ds − s(cd) 2 (c − d). We have f 0 (c) = 12 sc 2 g(c), where
g(c) = 2c
s+1
2
− (s + 1)cd
Since
0
g (c) = (s + 1) c
s−1
2
s−1
2
+ (s − 1)d
−d
s−1
2
s+1
2
.
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> 0,
g(c) is strictly increasing, g(c) > g(d) = 0, and hence f 0 (c) > 0. Therefore, f (c) is
strictly increasing, and then f (c) > f (d) = 0.
The right side of the inequality in (3.2) is equivalent to
a
(ab)a−b > 1.
b
Write this inequality as f (b) > 0, where
f (b) =
1+a−b
ln a − ln b.
1−a+b
In order to prove that f (b) > 0, it suffices to show that f 0 (b) < 0 for all b ∈ (0, a);
then f (b) is strictly decreasing, and hence f (b) > f (a) = 0. Since
f 0 (b) =
Inequalities With PowerExponential Functions
−2
1
ln a − ,
2
(1 − a + b)
b
the inequality f 0 (b) < 0 is equivalent to g(a) > 0, where
(1 − a + b)2
g(a) = 2 ln a +
.
b
vol. 10, iss. 1, art. 21, 2009
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Since 0 < b < a < 1, we have
g 0 (a) =
2 2(1 − a + b)
2(a − 1)(a − b)
−
=
< 0.
a
b
ab
Thus, g(a) is strictly decreasing on [b, 1], and therefore g(a) > g(1) = b > 0. This
completes the proof. Equality holds if and only if a = b.
Proof of Theorem 2.4. Without loss of generality, assume that a ≥ b. Let x = ra
and y = rb, where either x ≥ y ≥ 1 or 1 ≥ x ≥ y. The inequality (2.1) becomes
Inequalities With PowerExponential Functions
Vasile Cîrtoaje
vol. 10, iss. 1, art. 21, 2009
xx − y x ≥ rx−y (xy − y y ).
Since x ≥ y, xy − y y ≥ 0 and r ≤ e, it suffices to show that
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xx − y x ≥ ex−y (xy − y y ).
(3.3)
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For the nontrivial case x > y, using the substitutions c = xy and d = y y (where
c > d), we can write (3.3) as
x
y
x
y
c − d ≥ ex−y (c − d).
x
x
x−y
x
(xy) 2 > ex−y .
y
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x−y
x
(cd) 2y (c − d) > ex−y (c − d).
y
The left side of the inequality is just the left hand inequality in (3.2) for s =
the right side of the inequality is equivalent to
II
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In order to prove this inequality, we will show that
cy − dy >
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x
,
y
while
We write this inequality as f (x) > 0, where
1
f (x) = ln x − ln y + (x − y)(ln x + ln y) − x + y.
2
We have
f 0 (x) =
1 ln(xy)
y
1
+
−
−
x
2
2x 2
and
f 00 (x) =
x+y−2
.
2x2
Case x > y ≥ 1. Since f 00 (x) > 0, f 0 (x) is strictly increasing, and hence
f 0 (x) > f 0 (y) =
1
y
0
1
+ ln y − 1.
y
y−1
y2
0
Let g(y) = + ln y − 1. From g (y) =
> 0, it follows that g(y) is strictly
increasing, g(y) ≥ g(1) = 0, and hence f (x) > 0. Therefore, f (x) is strictly
increasing, and then f (x) > f (y) = 0.
Case 1 ≥ x > y. Since f 00 (x) < 0, f (x) is strictly concave on [y, 1]. Then, it suffices
to show that f (y) ≥ 0 and f (1) > 0. The first inequality is trivial, while the second
inequality is equivalent to g(y) > 0 for 0 < y < 1, where
2(y − 1)
g(y) =
− ln y.
y+1
From
g 0 (y) =
−(y − 1)2
< 0,
y(y + 1)2
Inequalities With PowerExponential Functions
Vasile Cîrtoaje
vol. 10, iss. 1, art. 21, 2009
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it follows that g(y) is strictly decreasing, and hence g(y) > g(1) = 0. This completes
the proof.
Equality holds if and only if a = b.
Proof of Theorem 2.5. (after an idea of Wolfgang Berndt [1]). We will show that
ara + brb < arb + bra
for r = (x + 1)e, a =
1
e
and b =
1
r
=
xex +
1
,
(x+1)e
where x > 0; that is
1
> x + 1.
(x + 1)x
Since ex > 1 + x, it suffices to prove that
1
> 1 − x2 .
(x + 1)x
For the nontrivial case 0 < x < 1, this inequality is equivalent to f (x) < 0, where
f (x) = ln (1 − x2 ) + x ln(x + 1).
We have
and
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vol. 10, iss. 1, art. 21, 2009
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x
f 0 (x) = ln(x + 1) −
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x(x − 3)
f 00 (x) =
.
(1 + x)(1 − x)2
Since f 00 (x) < 0, f 0 (x) is strictly decreasing for 0 < x < 1, and then f 0 (x) <
f 0 (0) = 0. Therefore, f (x) is strictly decreasing, and hence f (x) < f (0) = 0.
4.
Other Related Inequalities
Proposition 4.1. If a and b are positive real numbers such that min{a, b} ≤ 1, then
the inequality
(4.1)
a−ra + b−rb ≤ a−rb + b−ra
holds for any positive real number r.
Inequalities With PowerExponential Functions
Proof. Without loss of generality, assume that a ≤ b and a ≤ 1. From ar(b−a) ≤
−rb b−ra
br(b−a) we get b−rb ≤ a a−ra
, and
a−rb b−ra
− a−rb − b−ra
−ra
a
(a−ra − a−rb )(a−ra − b−ra )
=
≤ 0,
a−ra
a−ra + b−rb − a−rb − b−ra ≤ a−ra +
because b−ra ≤ a−ra ≤ a−rb .
Proposition 4.2. If a, b, c are positive real numbers, then
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vol. 10, iss. 1, art. 21, 2009
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(4.2)
aa + b b + c c ≥ ab + b c + c a .
This inequality, with a, b, c ∈ (0, 1), was posted as a conjecture on the Mathlinks
Forum by Zeikii [1].
Proof. Without loss of generality, assume that a = max{a, b, c}. There are three
cases to consider: a ≥ 1, c ≤ b ≤ a < 1 and b ≤ c ≤ a < 1.
Case a ≥ 1. By Theorem 2.3, we have bb + cc ≥ bc + cb . Thus, it suffices to prove
that
aa + c b ≥ ab + c a .
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For a = b, this inequality is an equality. Otherwise, for a > b, we substitute x = ab ,
y = cb and s = ab (where x ≥ 1, x ≥ y and s > 1) to rewrite the inequality as
f (x) ≥ 0, where
f (x) = xs − x − y s + y.
Since
f 0 (x) = sxs−1 − 1 ≥ s − 1 > 0,
f (x) is strictly increasing for x ≥ y, and therefore f (x) ≥ f (y) = 0.
Case c ≤ b ≤ a < 1. By Theorem 2.3, we have aa + bb ≥ ab + ba . Thus, it suffices
to show that
ba + c c ≥ bc + c a ,
which is equivalent to f (b) ≥ f (c), where f (x) = xa − xc . This inequality is true if
f 0 (x) ≥ 0 for c ≤ x ≤ b. From
Inequalities With PowerExponential Functions
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vol. 10, iss. 1, art. 21, 2009
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f 0 (x) = axa−1 − cxc−1
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= xc−1 (axa−c − c)
≥x
c−1
we need to show that a − c
inequality we have
(ac
a−c
1−a+c
− c) = x
c−1 a−c
c
(a − c
1−a+c
),
≥ 0. Since 0 < 1 − a + c ≤ 1, by Bernoulli’s
c1−a+c = (1 + (c − 1))1−a+c
≤ 1 + (1 − a + c)(c − 1) = a − c(a − c) ≤ a.
Case b ≤ c ≤ a < 1. The proof of this case is similar to the previous case. So the
proof is completed.
Equality holds if and only if a = b = c.
Conjecture 4.3. If a, b, c are positive real numbers, then
(4.3)
a2a + b2b + c2c ≥ a2b + b2c + c2a .
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Conjecture 4.4. Let r be a positive real number. The inequality
ara + brb + crc ≥ arb + brc + cra
(4.4)
holds for all positive real numbers a, b, c with a ≤ b ≤ c if and only if r ≤ e.
We can prove that the condition r ≤ e in Conjecture 4.4 is necessary by setting
c = b and applying Theorem 2.5.
Proposition 4.5. If a and b are nonnegative real numbers such that a + b = 2, then
2b
a +b
(4.5)
2a
≤ 2.
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Without loss of generality, assume that a ≥ b. Since 0 ≤ a − 1 < 1 and 0 < b ≤ 1,
by Bernoulli’s inequality we have
a ≤ 1 + b(a − 1) = 1 + b − b
and
2
ba = b · ba−1 ≤ b[1 + (a − 1)(b − 1)] = b2 (2 − b).
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Therefore,
2b
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vol. 10, iss. 1, art. 21, 2009
Proof. We will show the stronger inequality
2
a−b
2b
2a
a +b +
≤ 2.
2
b
Inequalities With PowerExponential Functions
2a
a +b +
a−b
2
2
− 2 ≤ (1 + b − b2 )2 + b4 (2 − b)2 + (1 − b)2 − 2
= b3 (b − 1)2 (b − 2) ≤ 0.
Close
Conjecture 4.6. Let r be a positive real number. The inequality
(4.6)
arb + bra ≤ 2
holds for all nonnegative real numbers a and b with a + b = 2 if and only if r ≤ 3.
Conjecture 4.7. If a and b are nonnegative real numbers such that a + b = 2, then
4
a−b
3b
3a
(4.7)
a +b +
≤ 2.
2
Inequalities With PowerExponential Functions
Vasile Cîrtoaje
vol. 10, iss. 1, art. 21, 2009
Conjecture 4.8. If a and b are nonnegative real numbers such that a + b = 1, then
(4.8)
a2b + b2a ≤ 1.
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References
[1] A. ZEIKII, V. CÎRTOAJE AND W. BERNDT, Mathlinks Forum, Nov.
2006, [ONLINE: http://www.mathlinks.ro/Forum/viewtopic.
php?t=118722].
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