Solution of the OQ. 2283

Solution of the OQ. 2283

R´obert Sz´asz and Aurel P´al Kup´an⁴¹

ABSTRACT.We determine the set ofα, for which the weighted H¨older mean Hα(a, b)is between theL(a, b)P´olya-Szeg˜o logarithmic and theI(a, b) exponential mean. Other results concerning this means are proved.

1. INTRODUCTION Let 0< a < b. The generalized H¨older mean is

Qα(a, b) = a^α1 +b^1α 2

!α

.

The P´olya &Szeg˝o logarithmic mean and the exponential mean are defined by L(a, b) = b−a

lnb−lna, I(a, b) = 1 e

b^b a^a

_b−¹_a .

In [4] the authors proved that, the inequality holds:

L(a, b)< I(a, b), a, b∈(0,∞).

Other results concerning these means were deduced in [1]. An exhaustive treatment of the topic can be found in [3].

The author of [2] proposed the following open question: determine the values α∈[2,∞) for which the inequalities hold

L(a, b)≤Q_α(a, b)≤I(a, b) for all a, b∈(0,∞), a < b.

The aim of this paper is to determine the desired set of α and to deduce some other inequalities concerning these means.

41Received: 02.02.2009

2000Mathematics Subject Classification. 26D99

Key words and phrases. Inequalities, logarithmic means, exponential means.

2. PRELIMINARIES We will need in our work the following results.

Lemma 1. Letα∈(0,2] be a fixed number.If s∈(¹₂,1) then the inequality 2^α(s^α(1−s) + (1−s)^αs)<1

holds.

Proof. Let g₁ : [¹₂,1)→Rbe the function defined by the equality:

g₁(s) = 2^α(s^α(1−s) + (1−s)^αs).

By differentiation,

g₁^′(s) = 2^α[αs(1−s)(s^α−2−(1−s)^α−2) + (1−s)^α−s^α].

Since s^α−2−(1−s)^α−2 <0 and (1−s)^α−s^α<0 for s∈(¹₂,1),shows g^′₁(s)<0 for all s∈(¹₂,1).

Thus the function g₁ is strictly decreasing and the inequality g₁(s)< g₁(1

2) = 1, s∈(1 2,1) follows.

Lemma 2. Ifs∈(¹₂,1) is a fixed number, then the function g2 : [2,3]→R defined by

g₂(α) = 2^α(s^α(1−s) + (1−s)^αs) is strictly increasing.

Proof. The expression ofg2 can be rewritten as follows:

g₂(α) = 2s(1−s)[(2s)^α−1+ 2(1−s)α−1

].

We get by differentiation:

g₂^′(α) = 2s(1−s)[(2s)^α−1ln 2s+ 2(1−s)α−1

ln 2(1−s) ].

Since 2s∈(1,2),2(1−s)∈(0,1) andα−1∈[1,2],follows that g₂^′(α)≥2s(1−s)[2sln 2s+ 2(1−s) ln 2(1−s)

].

On the other hand for the derivative of the function g₃: [¹₂,1)→R, g₃(s) = 2sln 2s+ 2(1−s) ln 2(1−s)

the following inequality holds:

g₃^′(s) = ln s

1−s >0, for all s∈(1 2,1).

Thereforeg3 is strictly increasing andg3(s)> g3(¹₂) = 0 for alls∈(¹₂,1).

Thusg₂^′(α)>0,for allα∈(2,3) and the assertion holds.

Lemma 3. [5]. II.2. Let n be a natural numbern≥2,and a_k∈(0,∞), k= 1, n be real numbers, not all equal. The functionh: (0,∞)→R defined by the equality:

h(α) = 1 n

Xn k=1

a

1 α

k

!α

is strictly decreasing.

3. THE MAIN RESULT

Theorem 1. Letα∈(0,3] be a fixed number. For alla, b∈(0,∞), a < b the following inequality holds:

L(a, b)< Qα(a, b). (1)

Proof. Let x= ^b_a,obviously x∈(1,∞).The inequality (1) is equivalent to x−1

lnx <

1 +x^α1 2

α

, x∈(1,∞).

We let s= ^xα1

1+x^α1 and this leads to the next equivalent form of (1):

αln s

1−s−[(2s)^α− 2(1−s)α

]>0, s∈(1

2,1). (2)

Consequently we have to study the function f₁ : [1

2,1)→R, f₁(s) =αln s

1−s−[(2s)^α− 2(1−s)α

].

We mark out two cases.

We assume first α∈[0,2].Since f₁^′(s) = α

s(1−s)[1−2^α s^α(1−s) +s(1−s)^α ],

Lemma 1 implies f₁^′(s)>0, s∈(¹₂,1).Thus f₁ is strictly increasing, and the desired inequality follows: f1(s)> f1(¹₂) = 0, s∈[¹₂,1).

The second case is α∈[2,3].

The equality f₁^′(s) = _s(1^α_−s)[1−g1(s)], s∈(¹₂,1) and Lemma 3 imply, that it is sufficient to prove

f₁^′(s)>0, for all s∈(1 2,1)

in case if α= 3,and then the inequality follows for everyα ∈[2,3].

In case ifα= 3,we have f₁^′(s) = 3

s(1−s)[1−2³ s³(1−s) +s(1−s)³ ] and

g₁(s) = 2³ s³(1−s) +s(1−s)³ .

Since g^′₁(s) = 8(1−2s)³<0, s∈(¹₂,1),it follows that g₁ is a decreasing mapping on (¹₂,1).Thus g1(s)< g1(¹₂) = 1 for alls∈(¹₂,1).Hence

f₁^′(s)>0, for all s∈(1

2,1) andα∈[2,3].

Consequentlyf₁(s)> f₁(¹₂) for s∈(¹₂,1),and (2) holds in this case too.

Remark 1. If α >3 then the inequality

L(a, b)≤Q_α(a, b) (1)

does not hold for every 0< a < b.

Proof. We have to prove that (2) does not hold provided α >3.Letg₁ be the function defined in the proof of Lemma 1. Sinceg₁^′′(¹₂) =α(α−3)>0 the continuity of g^′′₁ implies the existence of a real numberε >0 so that g^′′₁(s)>0 for all s∈[¹₂,¹₂ +ε).Hence g₁^′ is strictly increasing on [¹₂,¹₂ +ε).

Thusg₁^′(s)> g₁^′(¹₂) = 0 for alls∈(¹₂,¹₂ +ε).Thus we get thatg₁ is a strictly increasing mapping on [¹₂,¹₂ +ε) andg₁(s)> g₁(¹₂) = 1, s∈(¹₂,¹₂ +ε).This leads to

f₁^′(s)<0, s∈(1 2,1

2 +ε) and f₁(s)< f₁(1

2) = 0, s∈(1 2,1

2 +ε).

Hence (2) cannot be true for every s∈(¹₂,1).

Theorem 2. Ifα∈[³₂,∞) then the following inequality holds

Q_α(a, b)< I(a, b) for all a, b∈(0,∞), a < b. (3)

Proof. According to Lemma 3 we have to prove (3) only in case if α= ³₂. Using the notation x= ^b_a the inequality

a²³ +b²³ 2

³₂

< 1 e

b^b a^a

_b¹

−a

, a, b∈(0,∞), a < b is equivalent to

x

x−1lnx−3 2ln

1 +x²³ 2

−1>0, x∈(1,∞). (4) Let

f2 : [1,∞)→R, f2(x) = x

x−1lnx− 3 2ln

1 +x²³ 2

−1.

By differentiation

f₂^′(x) = 1 (x−1)²

−lnx+x−1−(x−1)² 1 x+x¹³

.

Let

u: [1∞)→R, u(x) =−lnx+x−1−(x−1)² 1 x+x¹³. Since

u^′(x) = (x−1)(x¹³ −1)³ 3x²³(x+x¹³)²

is positive for everyx∈(1,∞),it shows that u is strictly increasing on [1,∞).Therefore we haveu(x)> u(1) = 0, x∈(1,∞).Thus f₂^′(x) is positive for every x∈(1,∞) and so f₂ is strictly increasing on [1,∞) which means that the inequality

f₂(x)> f₂(1) = 0, x∈(1,∞) holds, and this is equivalent to (4).

Remark 2. If 0< α < ³₂ then the inequality

Qα(a, b)≤I(a, b) (5)

does not hold for all a, b∈(0,∞), a < b.

Proof. Inequality (5) is equivalent to:

xlnx−α(x−1) ln

1 +x^α1 2

−x+ 1≥0, x∈(1,∞). (6) Suppose 0< α < ³₂.The derivatives of the function f3: [1,∞)→R defined by

f₃(x) =xlnx−α(x−1) ln

1 +x^α1 2

−x+ 1 are:

f₃^′(x) = lnx−(x−1) x^α1−1

1 +x^α1 −αln

1 +x^1α 2

,

f₃^′′(x) = 1−x^α1

x(1 +x^α1)+ (1−x)(_α¹ −1)x^1α−2−x^α2−2 (1 +x^1α)² . Since

xlimց1

f₃^′′(x)

1−x = 3−2α 4α >0

there exists a positive real number ε >0,so that f₃^′′(x)<0, x∈(1,1 +ε).

Thusf₃^′ is decreasing on (1,1 +ε) and f₃^′(x)< f₃^′(1) = 0, x∈(1,1 +ε).

Thereforef₃ is also decreasing on (1,1 +ε) and it follows that

f3(x)< f3(1) = 0, x∈(1,1 +ε),and this inequality is in contradiction with (6).

Remark 3. If we denotex= _a^b,then the inequality

L(a, b)≥Q_α(a, b). a, b∈(0,∞), a < b (7) is equivalent to

1

lnx− 1 +x^α1 2(x−1)^α1

!α

≥0, x∈(1,∞).

But

xlim→∞

1

lnx − 1 +x^α1 2(x−1)^α1

!α

=−1

for everyα∈(0,∞) fixed number. This means that inequality (7) cannot be true for any α∈(0,∞).

Conclusions

1. The inequalities

L(a, b)< Qα(a, b)< I(a, b), hold for all a, b∈(0,∞), a < b if and only if α∈[³₂,3].

2. Remark 1 shows that there is no α∈(0,∞) so that

Q_α(a, b)≤L(a, b), for all a, b∈(0,∞), a < b.

Remark 4. A more general version of Theorem 1 and Theorem 2 can be found [6].

REFERENCES

[1] Anisiu Valeriu and Anisiu Mira Cristina, Refinement of Some Inequalities for Means, Revue D’Analyse et de Theorie de L’Approximation, Tome 35, No.1, 2006, pp.5-10

[2] Bencze Mih´aly, Open Question no 2283.Octogon Mathematical Magazine, Vol.14. No. 2. October 2006.

[3] Bullen P.S.,Handbook of Means and Their Inequalities,

Series:Mathematics and Its Applications, vol.560, 2nd ed., Kluwer Academic Publishers Group, Dordrecht, 2003

[4] Ivan, M. and Ra¸sa, I., Some Inequalities for Means, Seminar of Functional Equations, Approximation and Convexity, Cluj-Napoca, May 23-29,2000,pp.99-102

[5] P´olya, Gy.,Szeg˝o, G., Aufgaben und Lehrs¨atze aus der Analysis Springer Verlag. 1924

[6] Edward Neuman,A generalization of an inequality of JIA and CAU Journal of Inequalities in Pure and Applied Mathematics,Vol.5,Issue 1,