# Improvements of Logarithmic and Identric Mean Inequalities for Scalars and Operators

Authors: Aliaa Burqan; Abeer Abu-Snainah; Rania Saadeh
Journal: Journal of Applied Mathematics
 (2023)

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## Abstract

In this article, we provide refined inequalities for a convex Riemann’s integrable function using refinements of the classical Hermite-Hadamard inequality. The obtained results are applied on special functions to establish new improvements of inequalities on the weighted logarithmic mean and weighted identric mean. Moreover, corresponding operator inequalities are introduced based on the scalar inequalities and the monotonicity property for operators.


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## Body

## 1. Introduction

The theory of means has grown to occupy a prominent position in mathematics with hundreds of papers on the subject appearing every year. It has applications in so many diverse field, these include electrostatics, heat conduction, chemistry, and even medicine.The inequalities on means have attracted the attention of many mathematicians because of its implications. Pal et al. [1], introduced the weighted logarithmic mean and the weighted identric mean and presented the inequalities among weighted means as
(1)x#μy≤Lμx,y≤x∇μy,(2)x#μy≤Iμx,y≤x∇μy,where the weighted geometric mean x#μy=x1−μyμ, the weighted arithmetic mean x∇μy=1−μx+μy, the weighted logarithmic mean
(3)Lμx,y=1lny−lnx1−μμx1−μyμ−x+μ1−μy−x1−μyμ,and the weighted identric mean
(4)Iμx,y=1ex∇μy1−2μx∇μy/μ1−μy−xyμy/1−μx1−μx/μ1/y−x,for x,y>0 and μ∈0,1.The classical Hermite-Hadamard inequality, which was first discovered by Hermite in 1883 and independently proved in 1893 by Hadamard, involves a fundamental characteristic of convex functions and it states that iff is a convex real valued function on the interval x,y, then
(5)fx+y2≤1y−x∫xyftdt≤fx+fy2.As a generalization of the Hermite-Hadamard inequality, Pal et al. [1] produced the following new inequality
(6)fx∇μy≤Cf,μx,y≤fx∇μfy,where
(7)Cf,μx,y=∫01fx∇μtydt∇μ∫01f1−μy−xt+x∇μydt,for a convex Riemann’s integrable function f, x,y>0 and μ∈0,1. A refinement of inequality (6) was given by Furuichi and Minculete [2] and some interesting related integral inequalities can be found in [3], [4–6], and [7, 8].A motivating problem related to the Hermite-Hadamard inequality is the refinements of this inequality. One of these refinements, due to Feng [9], asserts that
(8)1y−x∫xyftdt≤12fx+fy2+fx+y2≤fx+fy2.On the other hand, Burqan [10] constructed a refinement of the first inequality in the Hermite-Hadamard inequality as follows:
(9)fx+y2≤12f3x+y4+fx+3y4≤1y−x∫xyftdt.In this work, we apply the above refinements of the Hermite-Hadamard inequality to obtain a refinement of the inequalities given in (6) that produces improvements of the inequalities in (1) and (2).


## 2. Main Results

In this section, we obtain the refined inequalities for (6) by using the refinements of the Hermite-Hadamard inequality given in (8) and (9).Theorem 1.
For every convex Riemann’s integrable functionf:x,y⟶ℝ and 0≤μ≤1, we have
(10)fx∇μy≤Af,μ1x,y≤Bf,μ1x,y≤Cf,μx,y≤Bf,μ2x,y≤Af,μ2x,y≤fx∇μfy,where
(11)Af,μ1x,y=fx∇μ/2y∇μfx∇1+μ/2y,Bf,μ1x,y=fx∇μ/4y∇fx∇3μ/4y∇μfx∇3μ+1/4y∇fx∇3+μ/4y,Cf,μx,y=∫01fx∇μtydt∇μ∫01f1−μy−xt+x∇μydt,Af,μ2x,y=fx∇μfy∇fx∇μy,Bf,μ2x,y=fx∇fx∇μy∇fx∇μ/2y∇μfx∇μy∇fy∇fx∇1+μ/2y.Proof.
Applying the refinements (8) and (9) of the Hermite-Hadamard inequality on two intervals x,1−μx+μy and 1−μx+μy,y, we obtain, respectively.
(12)f2−μx+μy2≤12f4−μx+μy4+f4−3μx+3μy4≤1μy−x∫x1−μx+μyftdt≤12fx+f1−μx+μy2+f2−μx+μy2≤fx+f1−μx+μy2,(13)f1−μx+1+μy2≤12f3−3μx+3μ+1y4+f1−μx+3+μy4≤11−μy−x∫1−μx+μyyftdt≤12f1−μx+μy+fy2+f1−μx+1+μy2≤f1−μx+μy+fy2.
Multiplying inequalities in (12) and (13) by 1−μ and μ, respectively, and summing each side, we obtain
(14)1−μf2−μx+μy2+μf1−μx+1+μy2≤1−μ2f4−μx+μy4+f4−3μx+3μy4+μ2f3−3μx+3μ+1y4+f1−μx+3+μy4≤1−μμy−x∫x1−μx+μyftdt+μ1−μy−x∫1−μx+μyyftdt≤1−μ2fx+f1−μx+μy2+f2−μx+μy2+μ2f1−μx+μy+fy2+f1−μx+1+μy2≤1−μfx+f1−μx+μy2+μf1−μx+μy+fy2.
Which is equivalent to(15)Af,μ1x,y≤Bf,μ1x,y≤1−μμy−x∫x1−μx+μyftdt+μ1−μy−x∫1−μx+μyyftdt≤Bf,μ2x,y≤Af,μ2x,y.
Now, by replacing the variables such ast=μy−xs+x in the first integral and t=1−μy−xu+1−μx+μy in the second integral given in (15), respectively, we have
(16)1−μμy−x∫x1−μx+μyftdt=1−μ∫01fμy−xs+xds=1−μ∫01fx∇μsyds.(17)μ1−μy−x∫1−μx+μyyftdt=μ∫01f1−μy−xu+1−μx+μydu=μ∫01f1−μy−xu+x∇μydu.
From (16) and (17) inequality, (15) becomes
(18)Af,μ1x,y≤Bf,μ1x,y≤Cf,μx,y≤Bf,μ2x,y≤Af,μ2x,y.
Finally, the proof is completed by noting that(19)fx∇μy=f2−2μx+2μy2=f1−μ2−μ+μ1−μx+μ1−μ+μ1+μy2≤1−μf2−μ2x+μ2y+μf1−μ2x+1+μ2y=Af,μ1x,y,Af,μ2x,y=fx∇μfy∇fx∇μy=1−μfx+μfy+f1−μx+μy2≤fx∇μfy+fx∇μfy2=fx∇μfy.In the following results, we get refinements of some inequalities that introduced by Raïssouli and Furuichi [11].Corollary 2.
Forx,y>0 and 0<μ<1, we have
(20)x#μy≤x#μ/2y∇μx#1+μ/2y≤x#μ/4y∇x#3μ/4y∇μx#3μ+1/4y∇x#3+μ/4y≤Lμx,y≤x#μy∇x∇x#μy∇μx#μy∇y∇x#1+μ/2y≤x∇μy∇x#μy≤x∇μy.Proof.
Applying Theorem1 on the convex function ft=et, we get for y≥x>0.
(21)e1−μx+μy≤1−μe2−μx+μy/2+μe1−μx+1+μy/2≤1−μ2e1−μ/4x+μ/4y+e1−3μ/4x+3μ/4y+μ2e1−3μ+1/4x+3μ+1/4y+e1−3+μ/4x+3+μ/4y≤1−μ∫01eμy−xt+xdt+μ∫01e1−μy−xt+μy+1−μxdt≤1−μ2ex+e1−μ+μy2+e1−μ/2x+μ/2y+μ2e1−μx+μy+ey2+e1−1+μ/2x+1+μ/2y≤1−μex+μey+e1−μx+μy2≤1−μex+μey.
By elementary calculations, we get(22)1−μ∫01eμy−xt+xdt+μ∫01e1−μy−xt+μy+1−μxdt=1y−x1−μμe1−μx+μy−ex+μ1−μey−e1−μx+μy.
Replacingex and ey with x and y, respectively, we obtain the inequalities (20) for y≥x>0 and μ∈0,1.
Dividingx in all sides of the obtaining inequalities and putting y/x=t≥1, we have
(23)tμ≤1−μtμ/2+μt1+μ/2≤1−μ2tμ/4+t3μ/4+μ2t3μ+1/4+t3+μ/4≤Lμ1,t≤1−μ2tμ+12+tμ/2+μ2t+tμ2+t1+μ/2≤121−μ+μt+tμ≤1−μ+μt.
Puttings=1/t≤1 and u=1−μ in (23) and then multiplying by s>0 to all sides to get
(24)su≤1−usu/2+us1+u/2≤u2su+3/4+s3u+1/4+1−u2s3u/4+su/4≤Lu1,s≤u2su+s2+su+1/2+1−u21+su2+su/2≤121−u+us+su≤1−u+us,for 0<s<1,u∈0,1.
By elementary calculations, we have the inequalities(25)tμ≤1−μtμ/2+μt1+μ/2≤1−μ2tμ/4+t4μ/4+μ2t3μ+1/4+t3+μ/4≤Lμ1,t≤1−μ2tμ+12+tμ/2+μ2t+tμ2+t1+μ/2≤121−μ+μt+tμ≤1−μ+μt,for t>0,μ∈0,1.Therefore, we complete the proof by puttingt=y/x for any x,y>0 in (25) and then multiplying all sides by x>0.Corollary 3.
Fory≥x>0 and 0<μ<1, we have
(26)x#μy≤x∇μy#x#μy≤x#x∇μy#x∇μ/2y#μx∇μy#y#x∇1+μ/2y1/2≤Iμx,y≤x∇μ/4y#x∇3μ/4y#μx∇3μ+1/4y#x∇3+μ/4y≤x∇μ/2y#μx∇1+μ/2y≤x∇μy.Proof.
Applying Theorem1 on the convex function ft=−logtt>0, we get for y≥x>0(27)1−μlogx+μlogy≤121−μlogx+μlogy+log1−μx+μy≤1−μ2logx+log1−μx+μy2+log1−μ2x+μ2y+μ2log1−μx+μy+logy2+log1−1+μ2x+1+μ2y≤1−μ∫01logμy−xt+xdt+μ∫01log1−μy−xt+1−μx+μydt≤1−μ2log1−μ4x+μ4y+log1−3μ4x+3μ4y+μ2log1−3μ+14x+3μ+14y+μ2log1−3+μ4x+3+μ4y≤1−μlog1−μ2x+μ2y+μlog1−1+μ2x+1+μ2y≤log1−μx+μy.
By elementary calculations, we have(28)logx1−μyμ≤logx1−μ/2yμ/21−μx+μy1/2≤logx1/21−μx+μy1/21−μ2x+μ2y1−μ/21−μx+μyμ/2yμ/21−1+μ2x+1+μ2yμ/2≤1−μ∫01logμy−xt+xdt+μ∫01log1−μy−xt+1−μx+μydt≤log1−μ4x+μ4y1−3μ4x+3μ4y1−μ/21−3μ+14x+3μ+14y1−3+μ4x+3+μ4yμ/2≤log1−μ2x+μ2y1−μ1−1+μ2x+1+μ2yμ≤log1−μx+μy.
Now, the calculations of the middle part of the above inequality (28) are given by
(29)1−μ∫01logμy−xt+xdt+μ∫01log1−μy−xt+1−μx+μydt=1−μμy−xμy−x+xlogμy−x+x−μy−x+x−xlogx+x+μ1−μy−x1−μy−x+1−μx+μylog1−μy−x+1−μx+μy−1−μy−x+1−μx+μy−1−μx+μylog1−μx+μy−1−μx+μy.=log1−μx+μy1−2μ1−μx+μy/μ1−μy−x+logx−x1−μ/μy−x+logyμy/1−μy−x+1−μμy−x−μy−x+μ1−μy−x−1−μy−x=log1e1−μx+μy1−2μ1−μx+μy/μ1−μy−xyμy/1−μx1−μx/μ1/y−x.
Substituting (28) in (29) and using the monotonicity function, we get the required result.
Our obtained results in the previous section can be extended to the operator inequalities, indicate the space of bounded linear operators on a the Hilbert spaceH by BH. For X,Y∈BH, we write X<Y to mean Y−X is positive definite, particularly, 0<X denotes that X is positive definite. The definitions of means in the scalar case can be used to raise these definitions to the operator level. For μ∈0,1 and positive definite operators X,Y∈BH, the weighted geometric operator mean and arithmetic operator mean are defined as follows:
(30)X#μY=X1/2X−1/2YX−1/2μX1/2,X∇μY=1−μX+μY.
It is known that an operator meanMX,Y is associated with the representing function ft=M1,t with a mean Mx,y for positive numbers x,y;(31)MX,Y=X1/2fX−1/2YX−1/2X1/2.
The weighted logarithmic operator meanLμX,Y is defined through the representing function FLμt=Lμ1,t for μ∈0,1(32)LμX,Y=X1/2FLμX−1/2YX−1/2X1/2,FLμt=Lμ1,t,and the weighted identric operator mean IμX,Y is defined through the representing function FIμt=Iμ1,t for μ∈0,1(33)IμX,Y=X1/2FIμX−1/2YX−1/2X1/2,FIμt=Iμ1,t.
Now, we are ready to present operator inequalities related to the logarithmic and identric means. The proofs are passed on the obtained scalar inequalities and the monotonicity property for operators which asserts that ifT∈BH is self-adjoint with a spectrum SpT and if f and g are continuous functions such that ft≥gt for all t∈SpT, then fT≥gT.Theorem 4.
LetX,Y∈BH be such that 0<X<Y and 0≤μ≤1. Then,
(34)X#μY≤X#μ/2Y∇μX#1+μ/2Y≤X#μ/4Y∇X#3μ/4Y∇μX#3μ+1/4Y∇X#3+μ/4Y≤LμX,Y≤X#μY∇X∇X#μY∇μX#μY∇Y∇X#1+μ/2Y≤X#μY∇X∇μY≤X∇μY.Proof.
LetT=X−1/2YX−1/2 and t∈σT. Then, we have T>I and so t>1. Corollary 2 implies that
(35)tμ≤1−μtμ/2+μt1+μ/2≤1−μ2tμ/4+t3μ/4+μ2t3μ+1/4+t3+μ/4≤Lμ1,t≤1−μ2tμ+12+tμ+μ2tμ+t2+t1+μ/2≤121−μ+μt+tμ≤1−μ+μt,where
(36)Lμ1,t=1lnt1−μμtμ−1+μ1−μt−tμ,t>0,0<μ<1.
By monotonicity property, we have(37)Tμ≤1−μTμ/2+μT1+μ/2≤1−μ2Tμ/4+T3μ/4+μ2T3μ+1/4+T3+μ/4≤LμI,T≤1−μ2Tμ+12+Tμ+μ2Tμ+T2+T1+μ/2≤12Tμ+1−μ+μT≤1−μ+μT.
By puttingT=X−1/2YX−1/2 and multiplying both sides by X1/2, we get
(38)X1/2X−1/2YX−1/2μX1/2≤1−μX1/2X−1/2YX−1/2μ/2X1/2+μX1/2X−1/2YX−1/21+μ/2X1/2≤1−μ2X1/2X−1/2YX−1/2μ/4+X−1/2YX−1/23μ/4X1/2+μ2X1/2X−1/2YX−1/23μ+1/4+X−1/2YX−1/23+μ/4X1/2≤X1/2LμI,X−1/2YX−1/2X1/2≤1−μ2X1/2X−1/2YX−1/2μ+I2+X−1/2YX−1/2μX1/2+μ2X1/2X−1/2YX−1/2μ+X−1/2YX−1/22+X−1/2YX−1/21+μ/2X1/2≤12X1/2X−1/2YX−1/2μX1/2+X1/21−μX1/2+μX1/2X−1/2YX−1/2X1/2≤X1/21−μX1/2+μX1/2X−1/2YX−1/2X1/2.
In the following theorem, we introduce new operator inequalities related to the identric mean.Theorem 5.
LetX,Y∈BH be such that 0<X<Y and 0≤μ≤1. Then,
(39)X#μY≤X∇μY#X#μY≤X#X∇μY#X∇μ/2Y#μX#μY#Y#X∇1+μ/2Y≤IμX,Y≤X∇μ/4Y#X∇3μ/4Y#μX∇3μ+1/4Y#X∇3+μ/4Y≤X∇μ/2Y#μX∇1+μ/2Y≤X∇μY.Proof.
LetT=X−1/2YX−1/2 and t∈σT. Then, we have T>I and so t>1. Corollary 3 implies that
(40)tμ≤1−μtμ/2+μt1+μ/2≤1−μ2tμ/4+t3μ/4+μ2t3μ+1/4+t3+μ/4≤Lμ1,t≤1−μ2tμ+12+tμ+μ2tμ+t2+t1+μ/2≤121−μ+μt+tμ≤1−μ+μt,where
(41)Iμ1,t=log1e1−μ+μt1−2μ1−μ+μt/μ1−μt−1μy/t1−μ1−μ/1μ1/y−x,t>0,0<μ<1.
By monotonicity property, we have(42)Tμ≤1−μ+μTTμ1/2≤1−μ+μT1/21−μ2+μ2T1−μ/21−μ+μTT1/21−μ2+1+μ2Tμ/2≤Iμ1,T≤1−μ4+μ4T1−3μ4+3μ4T1−μ/23−3μ4+3μ+14T1−μ4+3+μ4Tμ/2≤1−μ2+μ2T1−μ1−μ2+1+μ2Tμ≤1−μ+μT.
By puttingT=X−1/2YX−1/2 and multiplying both sides by X1/2. After simple computations, we get
(43)X#μY≤X∇μY#X#μY≤X#X∇μY#X∇μ/2Y#μX#μY#Y#X∇1+μ/2Y≤IμX,Y≤X∇μ/4Y#X∇3μ/4Y#μX∇3μ+1/4Y#X∇3+μ/4Y≤X∇μ/2Y#μX∇1+μ/2Y≤X∇μY.
Thus, the proof of Theorem5 is completed.


## 3. Conclusions

Mean functions are effective tools to obtain new approaches to inequalities and give estimations in numerical integration, statistics, and also real world problems. The concept of inequalities that will shed light on some unknown points about approximation theory and convex programming is a novelty in applied sciences as well as in mathematics. In addition, it has increased its contribution to the literature with its applications in areas such as engineering, statistics, and information theory. In this research, refined inequalities for a convex Riemann’s integrable function using refinements of the classical Hermite-Hadamard were introduced and implemented to get improvements of the weighted logarithmic and identric mean inequalities. In recent years, progress and achievements in this field have been very rich; we hope this research is one of these achievements that may open new doors for researchers to work on convex Riemann’s integrable functions in different directions.



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