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volume 5, issue 2, article 32, 2004.

Received 23 September, 2003;

accepted 01 March, 2004.

Communicated by:J.E. Peˇcari´c

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Journal of Inequalities in Pure and Applied Mathematics

A SUFFICIENT CONDITION FOR THE INTEGRAL VERSION OF MARTINS’ INEQUALITY

VANIA MASCIONI

Department of Mathematical Sciences Ball State University

Muncie, IN 47306-0490, USA.

EMail:vdm@cs.bsu.edu

URL:http://www.cs.bsu.edu/homepages/vdm/

c

2000Victoria University ISSN (electronic): 1443-5756 126-03

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A Sufficient Condition for the Integral Version of Martins’

Inequality Vania Mascioni

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J. Ineq. Pure and Appl. Math. 5(2) Art. 32, 2004

http://jipam.vu.edu.au

Recently a number of papers have appeared on Martins’ inequality:

(1) 1

n

X

i=1

i^r

, 1 n+ 1

n+1

X

i=1

i^r

!¹_r

<

√n

n!

n+1p

(n+ 1)!,

which holds forr >0andn∈N(see [2]). For example, in [1] it is proved that

(2) 1

n

X

i=1

a^r_i

, 1 n+ 1

n+1

X

i=1

a^r_i

!¹_r

<

√n

a_n!

n+1p a_n+1!,

where{a_i}is an increasing non-constant sequence of positive numbers satisfy- ing (1) a`/a`+1 ≥ a`−1/a` and (2) (a`+1/a`)^` ≥ (a`/a`−1)^`−1 for ` > 1 (and where it is agreed that a_n! stands forQn

i=1a_i). In particular, the authors show that the sequencea_i =ci+dgives a generalization of Martins’ result whenever c > 0andd≥0.

On a parallel path, continuous versions of the inequality have been investi- gated, and in [4] F. Qi and B.-N. Guo ask under which conditions the following holds:

(3)

1 b−a

Rb

a f^r(x)dx

1 b+δ−a

Rb+δ

a f^r(x)dx

!¹_r

≤

exp

1 b−a

Rb

a lnf(x)dx exp

1 b+δ−a

Rb+δ

a lnf(x)dx ,

whenever f is a positive, increasing and integrable function on the closed in- terval [a, b+δ] (withb > a and δ > 0) and r > 0is arbitrary. In a related

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result, in [3] N. Towghi and F. Qi prove that for allr >0and any non-negative, integrablef we have

(4) sup_x∈[a,b]f(x)

sup_x∈[a,b+δ]f(x) ≤

1 b−a

Rb

af^r(x)dx

1 b+δ−a

Rb+δ

a f^r(x)dx

!¹_r

(note that the l.h.s. in (4) is the limit forr→ ∞of the r.h.s.). In another remark, they note that (3) itself fails without extra assumptions. The issue, then, is at least to identify a sufficient hypothesis, and this is the aim of the present paper.

While logarithmic convexity of f has been identified as sufficient in related inequalities, our result below requires a strictly weaker hypothesis:

Theorem 1. Let f be a nondecreasing, positive, twice differentiable function onR⁺such that

(5) t(lnf(t))⁰⁰+ (lnf(t))⁰ ≥0 for allt >0. Then

(6) F(t) :=

1 t−a

Rt

af(x)dx exp

1 t−a

Rt

alnf(x)dx

is non-decreasing on[a,∞]for everya ≥ 0and therefore inequality (3) holds forf and every choice of0≤a < b, andr, δ >0.

Proof. It is plain that iffsatisfies (5) thenf^ralso does (for everyr >0), and so the last statement is a trivial consequence of functionF being non-decreasing.

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Fixa ≥0and in the following always assume thatt ≥a. Note that condition (5) implies

(7) (t−a)(lnf(t))⁰⁰+ (lnf(t))⁰ ≥0

for all t ≥ a (we are assuming that f(t) is non-decreasing, and therefore (lnf(t))⁰ ≥0). Computing the derivatives in (7) gives

(8) (t−a)f(t)f⁰⁰(t)−(f⁰(t))²

f²(t) +f⁰(t) f(t) ≥0 which is in turn equivalent to

(9)

− (t−a)f(t)

1 + (t−a)f⁰(t)/f(t)+ Z t

a

f(x)dx ⁰

≥0

(if you apply the quotient rule to differentiate the first summand in (9), and collect the l.h.s. over the common denominator, then the numerator is seen to be(t−a)f(t)times the l.h.s. in (8)). Now, (9) implies

(10) (t−a)f(t)

1 + (t−a)f⁰(t)/f(t) ≤ Z t

a

f(x)dx≤(t−a)f(t),

where the second inequality is due tof⁰ being non-decreasing. Next, consider- ing the left hand side of the following inequality as a quadratic polynomial in Rt

af(x)dx, (10) is seen to be equivalent to (11)

Z t

a

f(x)dx ²

1 + (t−a)f⁰(t) f(t)

− Z t

a

f(x)dx

2(t−a)f(t) + (t−a)²f⁰(t)

+ (t−a)²f²(t)≤0

(5)

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(inequality (11) says that Rt

af(x)dxmust lie between the two solutions of the quadratic polynomial, and the quadratic formula says that these two solutions are the l.h.s. and the r.h.s. of (10)).

Dividing both sides of (11) by Rt

af(x)dx2

and rearranging the terms we then obtain the equivalent form

(12) (t−a)²f(t) Rt

af(x)dx

!0

≥

(t−a) + (t−a) lnf(t)− Z t

a

lnf(x)dx ⁰

,

which clearly implies (13) (t−a)²f(t)

Rt

af(x)dx ≥(t−a) + (t−a) lnf(t)− Z t

a

lnf(x)dx

(since both sides vanish whent =a). Finally, if we divide (13) by(t−a)²we obtain

(14) f(t)

Rt

af(x)dx − 1

t−a − 1

t−alnf(t) + 1 (t−a)²

Z t

a

lnf(x)dx ≥0,

which is equivalent to (15)

ln

Z t

a

f(x)dx−ln(t−a)− 1 t−a

Z t

a

lnf(x)dx ⁰

≥0.

But this amounts to saying that the derivative of the natural logarithm of (16)

1 t−a

Rt

af(x)dx exp

1 t−a

Rt

alnf(x)dx

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is non-negative: the latter function oftmust therefore be non-increasing. Suffi- ciency of condition (5) is thus proved.

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References

[1] T.H. CHAN, P. GAOANDF. QI, On a generalization of Martins’ inequality, Monatsh. Math., 138 (2003) 179–87.

[2] J.S. MARTINS, Arithmetic and geometric means, an application to Lorentz sequence spaces. Math. Nachr., 139 (1988), 281–88.

[3] N. TOWGHI AND F. QI, An inequality for the ratios of the arithmetic means of functions with a positive parameter, RGMIA Res. Rep. Coll., 4(2) (2001), Art. 15 (electronic). ONLINE [http://rgmia.vu.edu.au/

v4n2.html].

[4] F. QI AND B.N. GUO, An algebraic inequality II, RGMIA Res. Rep. Coll., 4(1) (2001), Art. 8 (electronic). ONLINE [http://rgmia.vu.edu.

au/v4n1.html].

[5] F. QI AND B.N. GUO, An inequality between ratio of the extended loga- rithmic means and ratio of the exponential means, Taiwanese J. Math., 7 (2003), 229–37.