The monotonicity of the ratio between generalized logarithmic means

J. Math. Anal. Appl. 345 (2008) 86–89

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The monotonicity of the ratio between generalized logarithmic means ✩ Chao-Ping Chen College of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan 454010, China

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 30 September 2007 Available online 8 April 2008 Submitted by B.S. Thomson

A monotonicity result for the ratio between two generalized logarithmic means is established. As an application, an inequality of Alzer for negative powers is extended to all real numbers. © 2008 Elsevier Inc. All rights reserved.

Keywords: Monotonicity Generalized logarithmic mean Inequality

1. Introduction It was proved in [1,11,14] that n n+1

 <

n 1

n

r

i /

i =1

n +1 

1 n+1

1/r i

r

(1)

i =1

for all natural numbers n, and all real r > 0, and the lower bound is the best possible. (1) is called Alzer’s inequality. In [2,3] Alzer’s inequality was extended to all real r. Alzer’s inequality was generalized in [4–7,9,10,15]. The generalized logarithmic mean L r (a, b) of two positive numbers a and b is defined in [12,13] for a = b by L r (a, b) = a and for a = b by



L r (a, b) =

br +1 − ar +1

1/r

, r = −1, 0, (r + 1)(b − a) b−a L −1 (a, b) = = L (a, b), ln b − ln a  b 1/(b−a) 1 b L 0 (a, b) = = I (a, b). a e

a

Clearly, L 1 (a, b) = A (a, b),

L −2 (a, b) = G (a, b),

where A, G, L, I are arithmetic, geometric, logarithmic and identric means, respectively. It is known that L r (a, b) for a = b is a strictly increasing function of r. Clearly, for a = b, G (a, b) < L (a, b) < I (a, b) < A (a, b). ✩

(2)

This work was supported by the Natural Science Research Project of Education Department of Henan Province (2007110011). E-mail addresses: [email protected], [email protected].

0022-247X/$ – see front matter doi:10.1016/j.jmaa.2008.03.071

© 2008 Elsevier Inc.

All rights reserved.

C.-P. Chen / J. Math. Anal. Appl. 345 (2008) 86–89

87

In particular, lim L r (a, b) = min{a, b},

lim L r (a, b) = max{a, b}.

r →−∞

(3)

r →+∞

In this paper, we shall establish a monotonicity property for the ratio between two generalized logarithmic means. As an application, we extend Alzer’s inequality to all real r. Theorem 1. Let b, δ and ε be positive numbers. Then the function f (r ) =

L r (b, b + ε ) L r (b + δ, b + δ + ε )

is strictly increasing in r ∈ (−∞, +∞). Remark 1. By Theorem 1, we obtain the following result: Let b, δ and b b+δ

L r (b, b + ε )

<

L r (b + δ, b + δ + ε )

<

b+ε b+δ+ε

ε be positive numbers. Then for r ∈ R, (4)

.

The lower and upper bounds in (4) are the best possible, since lim

r →−∞

lim

r →+∞

L r (b, b + δ) L r (b + δ, b + δ + ε ) L r (b, b + δ) L r (b + δ, b + δ + ε )

b

=

,

b+δ b+δ

=

b+δ+ε

.

Theorem 2. Let n be a natural number. Then, for all real r, n n+1

 <

n 1

n

i /

i =1

n +1 

1

r

n+1

1/r i

r

< 1.

(5)

i =1

n The lower bound n+ and the upper bound 1 are the best possible. 1

Remark 2. (i) The right hand inequality of (5) is equivalent to 1 n+1

n +1 

ir ≷

i =1

n 1

n

ir ,

according as r ≷ 0,

(6)

i =1

increases with n according as r ≷ 0. decreases (ii) The left hand inequality of (5) is equivalent to

i.e. the arithmetic mean of the rth powers of 1 to n

1 n+1

n +1  i =1

r

i n+1

i.e. the Riemann sum for

 r

≶ 1 0

n 1

i

n

n

i =1

,

according as r ≷ 0,

xr dx with n equal subintervals

(7)

decreases with n according as r ≷ 0. increases

2. Proofs of Theorems 1 and 2 Proof of Theorem 1. Define for r ∈ (−∞, ∞),

⎧ ⎨ ln(

(b+ε )r +1 −br +1 ), (b+δ+ε )r +1 −(b+δ)r +1 ϕ (r ) = ⎩ ln( ln[(b+ε)/b] ), ln[(b+δ+ε )/(b+δ)]

Then, ln f (r ) =

 ϕ (r ) r

,

r = −1, r = −1.

r = 0,

ϕ  (0), r = 0.

In order to prove that ln f (r ) is strictly increasing, it suffices to show that tation reveals that

ϕ (−1 − r ) = ϕ (−1 + r ) + r ln

(b + δ + ε )(b + δ) , b(b + ε )

ϕ is strictly convex on (−∞, +∞). Easy compu-

88

C.-P. Chen / J. Math. Anal. Appl. 345 (2008) 86–89

which implies that ϕ  (−1 − r ) = ϕ  (−1 + r ), and then ϕ (r ) has the same convexity on both (−∞, −1) and (−1, +∞). Hence, it is sufficient to prove that ϕ is strictly convex on (−1, +∞). A simple computation yields b+δ r +1 b+δ r +1 2 ( b+δ+ [ln( b+δ+ ] ε) ε)

(r + 1)2 ϕ  (r ) =

−

b+δ r +1 2 [1 − ( b+δ+ ] ε)

( b+b ε )r +1 [ln( b+b ε )r +1 ]2 [1 − ( b+b ε )r +1 ]2

.

Define for 0 < t < 1,

ω(t ) =

t (ln t )2

(1 − t )2

.

Easy calculation gives



∞ 

ω (t ) n−1 = (1 + t ) ln t + 2(1 − t ) = − (1 − t )n+1 < 0, (1 − t )t ln t · ω(t ) n(n + 1) n=2

ω (t ) > 0 and ω(t ) is increasing for 0 < t < 1. Since b, δ and ε are positive numbers, then, for r > −1,

which means that

 0<

r +1

b b+ε

Hence, we obtain



<

r +1

b+δ

.

b+δ+ε

ϕ  (r ) > 0 for r > −1, and then ϕ (r ) is strictly convex on (−1, ∞). The proof of Theorem 1 is complete. 2

Proof of Theorem 2. For r = 0 (5) can be interpreted as

√ n

n n+1

<

√ n +1

n! < 1, (n + 1)!

see [8], because of

 lim

r →0

n 1

n

i =1

r

i /

1 n+1

n +1 

1/r i

r

√ n =

i =1

√

n +1

n!

(n + 1)!

.

Since (n + 1)r ≷ all of {1r , 2r , . . . , nr } according as r ≷ 0, (n + 1)r ≷ r ≷ 0. This proves (6).

n+1

n r +1 r From (7), ( n+ ) i =1 i ≶ 1 according as r + 1 ≷ 0), n 

ir ≷

i =1

nr +1 (n + 1)r

(n + 1)r +1 − nr +1

n

i =1 i

,

r

1 n

n

i =1 i

r

, so

r +1

n r +1 , according as r ≷ 0, i.e. nn+1 ≶ [1 − ( n+ ) ] 1

1 n+1

n

n+1

i =1 i

i =1

r

ir ≷

1 n

n

i =1 i

r

, according as

, i.e. (since (n + 1)r +1 − nr +1 ≷ 0

according as r (r + 1) ≷ 0.

(8)

n+1

n

−1 > −1 , so (7) is true when Our argument says nothing when r = −1. However, when r = −1, from (7) we get i =1 i i =1 i r = −1. Evidently, our ‘general’ argument deals with all the non-obvious cases of the theorem, but fails in this special case, in which the theorem is obvious. We may thus assume r (r + 1) ≷ 0 in all other cases to be proved. r Now we prove (8) by mathematical induction. Our induction hypothesis T (n) is then (8). T (1) is 1 ≷ 2r +21 −1 , according

as r (r + 1) ≷ 0. But 2r +1 − 1 ≷ 0, according as r + 1 ≷ 0. Thus T (1) is 2r +1 − 1 ≷ 2r , according as r (r + 1)/(r + 1) = r ≷ 0, i.e. T (1) is 2r ≷ 1, according as r ≷ 0, which is true. T (n) implies n +1 

ir ≷

i =1

nr +1 (n + 1)r

(n + 1)r +1 − nr +1

+ (n + 1)r =

(n + 1)2r +1 , (n + 1)r +1 − nr +1

according as r (r + 1) ≷ 0. This implies T (n + 1) if

(n + 1)r +1 (n + 2)r (n + 1)2r +1 ≷ , r + 1 r + 1 (n + 1) −n (n + 2)r +1 − (n + 1)r +1 according as r (r + 1) ≷ 0, i.e. (since (n + 1)r +1 − nr +1 ≷ 0 according as r + 1 ≷ 0),

(n + 1)r +1 − nr +1 (n + 1)r ≷ , r (n + 2) (n + 2)r +1 − (n + 1)r +1

(9)

according as r (r + 1)/(r + 1) = r ≷ 0. (9) can be written as n+1 n+2



>

(n + 1)r +1 − nr +1 (n + 2)r +1 − (n + 1)r +1

1/r

=

L r (n, n + 1) L r (n + 1, n + 2)

,

C.-P. Chen / J. Math. Anal. Appl. 345 (2008) 86–89

89

which is true by the right inequality of (4). This proves (7). It is easy to see that

 lim

r →+∞

n

 lim

r →−∞

n 1 i =1

n 1

n

ir /

i =1

r

i /

1 n+1 1 n+1

n+1 

1/r ir

i =1 n +1 

=

n n+1

(10)

,

1/r i

r

= 1.

(11)

i =1

Thus, both bounds in (5) are the best possible. The proof of Theorem 2 is complete.

2

References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]

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