Volume 10 (2009), Issue 3, Article 72, 5 pp.
SOLUTION OF ONE CONJECTURE ON INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS ˇ LADISLAV MATEJÍCKA I NSTITUTE OF I NFORMATION E NGINEERING , AUTOMATION AND M ATHEMATICS FACULTY OF C HEMICAL F OOD T ECHNOLOGY S LOVAK U NIVERSITY OF T ECHNOLOGY IN B RATISLAVA S LOVAKIA
[email protected]
Received 20 July, 2009; accepted 24 August, 2009 Communicated by S.S. Dragomir
A BSTRACT. 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]. Key words and phrases: Inequality, Power-exponential functions. 2000 Mathematics Subject Classification. 26D10.
1. I NTRODUCTION 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 arb + bra ≤ 2
(1.1)
holds for all nonnegative real numbers a and b with a + b = 2, if and only if r ≤ 3. 2. P ROOF OF C ONJECTURE 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,
1 2− x
3
3x(2− x1 ) 6x−3 1 12 6 1 1 + =8− + 2− 3+ x x x x x
The author is deeply grateful to Professor Vasile Cîrtoaje for his valuable remarks, suggestions and for his improving some inequalities in the paper. 193-09
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6x−3 and if we show that x1 > −6 + 12 − x62 + x13 then the inequality (2.1) will be fulfilled for x all x > 1. Put t = x1 , then 0 < t < 1. The inequality (2.1) becomes 6
t t > t3 (t3 − 6t2 + 12t − 6) = t3 β(t), 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 0 ≤ t ≤ t0 . Thus, 6 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 From α(1) = 0, it suffices to show that α0 (t) < 0 for t0 < t < 1, where 6 6 1 3t2 − 12t + 12 0 −3 − 3 . α (t) = − 2 ln t + t t t t − 6t2 + 12t − 6 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) 2 + +1 (t3 − 6t2 + 12t − 6)2 t 6 5 4 3 2 t − 12t + 56t − 120t + 120t − 48t 2 + + 1. = (t3 − 6t2 + 12t − 6)2 t
γ 0 (t) =
γ 0 (t) < 0 is equivalent to p(t) = 2t7 − 22t6 + 92t5 − 156t4 + 24t3 + 240t2 − 252t + 72 < 0. From p(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 00 (t) < 0 and (2.2) will be proved. Indeed, for t0 < t < 1, we have q 00 (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. 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
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 72, 5 pp.
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S OLUTION O F O NE C ONJECTURE
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The inequality ϕ0 (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 2 If δ(x) = 1−x − ln(1 + x) ≤ 0, then (2.4) is evident. Since δ 0 (x) = − (1+x) 2 − 1+x < 0 1+x 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 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
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 (2.7) 6x2 − 6x4 − (9x4 + 13x3 + 5x2 + 7x + 6)A − (9x4 − 13x3 + 5x2 − 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
A2 =
∞ X 2(−1)n+1 n=1
n+1
x
n X 1
n+1
,
B=−
∞ X n=0
!
1 xn+1 , n+1
∞ X
n
X1 2 xn+1 , B 2 = n+1 i n + 1 i=1 i n=1 i=1 ! ∞ 2n+1 X X (−1)i+1 1 AB = − x2n+2 . n + 1 i=1 i n=0
Since A2 + B 2 =
4 n+1 n=1,3,5,... X
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 72, 5 pp.
n X 1 i=1
i
! xn+1 ,
! xn+1
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and 4 n+1
n X 1 i=1
!
i
4 ≤ n+1
n−1 1+ 2
= 2,
we have 4 A2 + B 2 = 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
! xn+1
2x8 11 4 137 6 = 2x + x + x + . 6 90 1 − x2 From this and from the previous Taylor’s formulas we have x8 1 4 1 6 1 2 , (2.8) A + B > −x − x − x − 2 3 4 1 − x2 2
(2.9)
(2.10)
(2.11)
2 2 2 A − B > 2x + x3 + x5 + x7 , 3 5 7 A2 + B 2 < 2x2 +
11 4 137 6 2x8 x + x + , 6 90 1 − x2
5 A2 − B 2 < −2x3 − x5 , 3
5 4 x for 0 < x < 1. 12 Now, having in view (2.12) and the obvious inequality A + B < 0, to prove (2.7) it suffices to show that (2.12)
AB < −x2 −
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 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, 12 12
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 72, 5 pp.
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S OLUTION O F O NE C ONJECTURE
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which can be rewritten as 35 377 6 19 8 137 10 (2.13) − x4 + x + x − x + 6(x8 + x10 ) 2 30 2 30 55 4 1 6 5 8 2 − (A + B) 6 + 2x + x − x − x + (7x + 13x3 )(B − A) ≤ 0. 4 2 4 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 8 x 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
