# Optimal Inequalities between Harmonic, Geometric, Logarithmic, and Arithmetic-Geometric Means

Authors: Yu-Ming Chu; Miao-Kun Wang
Journal: Journal of Applied Mathematics
 (2011)

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## Abstract

We find the least valuesp, q, and s in (0, 1/2) such that the
inequalities H(pa+(1−p)b, pb+(1−p)a)>AG(a,b), G(qa+(1−q)b,qb+(1−q)a)>AG(a,b), and L(sa+(1−s)b,sb+(1−s)a)>AG(a,b) hold for all
a,b>0 with a≠b, respectively. Here AG(a,b),H(a,b),G(a,b), and L(a,b) denote the arithmetic-geometric, harmonic, geometric, and logarithmic means
of two positive numbers a and b, respectively.


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## Body

## 1. Introduction

The classical arithmetic-geometric meanAG(a,b) of two positive real numbers a and b is defined as the common limit of sequences {an} and {bn}, which are given by(1.1)a0=a,b0=b,an+1=an+bn2,bn+1=anbn.LetH(a,b)=2ab/(a+b), G(a,b)=ab, L(a,b)=(a-b)/(loga-logb),I(a,b)=(1/e)(bb/aa)1/(b-a), A(a,b)=(a+b)/2, and Mp(a,b)=[(ap+bp)/2]1/p(p≠0) and M0(a,b)=ab be the harmonic, geometric, logarithmic, identric, arithmetic, and p-th power means of two positive numbers a and b with a≠b, respectively. Then it is well known that (1.2)min⁡{a,b}<H(a,b)=M-1(a,b)<G(a,b)=M0(a,b)<L(a,b)<I(a,b)<A(a,b)=M-1(a,b)<max⁡{a,b}
for all a,b>0 with a≠b.Recently, the inequalities for means have been the subject of intensive research. In particular, many remarkable inequalities for arithmetic-geometric mean can be found in the literature [1–9].Carlson and Vuorinen [2], and Bracken [9] proved that (1.3)L(a,b)<AG(a,b)
for all a,b>0 with a≠b.In [3], Vamanamurthy and Vuorinen established the following inequalities: (1.4)AG(a,b)<I(a,b)<A(a,b),AG(a,b)<M1/2(a,b),AG(a,b)<π2L(a,b),M1(a,b)<AG(a2,b2)AG(a,b)<M2(a,b)
for all a,b>0 with a≠b.We recall the Gauss identity [6, 7](1.5)AG(1,r′)K(r)=π2
for r∈[0,1) and r′=1-r2. As usual, 𝒦 and ℰ denote the complete elliptic integrals [8] given by(1.6)K(r)=∫0π/2(1-r2sin⁡2θ)-1/2dθ=π2∑n=0∞(1/2,n)2(n!)2r2n,K′(r)=K(r′),E(r)=∫0π/2(1-r2sin⁡2θ)1/2dθ=π2∑n=0∞(-1/2,n)(1/2,n)(n!)2r2n,E′(r)=E(r′),where(a,0)=1 for a≠0, and (a,n)=∏k=0n-1(a+k).For fixeda,b>0 with a≠b and x∈[0,1/2], let(1.7)f1(x)=H(xa+(1-x)b,xb+(1-x)a),(1.8)f2(x)=G(xa+(1-x)b,xb+(1-x)a),(1.9)f3(x)=L(xa+(1-x)b,xb+(1-x)a).Then it is not difficult to verify thatf1(x),f2(x), and f3(x) are continuous and strictly increasing in [0,1/2], respectively. Note that f1(0)=H(a,b)<AG(a,b)<f1(1/2)=A(a,b), f2(0)=G(a,b)<AG(a,b)<f2(1/2) = A(a,b) and f3(0)=L(a,b)<AG(a,b)<f3(1/2)=A(a,b).Therefore, it is natural to ask what are the least valuesp, q, and s in (0,1/2) such that the inequalities H(pa+(1-p)b,pb+(1-p)a)>AG(a,b), G(qa+(1-q)b,qb + (1-q)a)>AG(a,b), and L(sa+(1-s)b,sb+(1-s)a)>AG(a,b) hold for all a,b>0 with a≠b, respectively. The main purpose of this paper is to answer these questions. Our main results are Theorems 1.1–1.3.Theorem 1.1.
Ifp∈(0,1/2), then inequality
(1.10)H(pa+(1-p)b,pb+(1-p)a)>AG(a,b)
holds for all a,b>0 with a≠b if and only if p≥1/4.Theorem 1.2.
Ifq∈(0,1/2), then inequality
(1.11)G(qa+(1-q)b,qb+(1-q)a)>AG(a,b)
holds for all a,b>0 with a≠b if and only if q≥1/2-2/4.Theorem 1.3.
Ifs∈(0,1/2), then inequality
(1.12)L(sa+(1-s)b,sb+(1-s)a)>AG(a,b)
holds for all a,b>0 with a≠b if and only if s≥1/2-3/4.


## 2. Lemmas

In order to establish our main results we need several formulas and lemmas, which we present in this section.For0<r<1, the following derivative formulas were presented in [6, Appendix E, pp. 474-475]: (2.1)dKdr=E-r′2Krr′2,dEdr=E-Kr,d(E-r′2K)dr=rK,d(K-E)dr=rEr′2,(2.2)K(2r1+r)=(1+r)K(r).The following Lemma2.1 can be found in [6, Theorem 3.21(7) and Exercise 3.43(4)].Lemma 2.1.
(1)  (1+r′2)ℰ(r)-2r′2𝒦(r) is strictly increasing from (0,1) onto (0,1);
(2)  ℰ(r)/r′1/2 is strictly increasing from (0,1) onto (π/2,+∞).Lemma 2.2.
Inequality(2.3)2πK(r)1-12r2>1
holds for all r∈(0,1).Proof.
Let(2.4)f(r)=log⁡[2πK(r)1-12r2].
Then simple computations lead to
(2.5)f(0)=0,(2.6)f′(r)=E(r)-r′2K(r)rr′2K(r)-r2-r2=(1+r′2)E(r)-2r′2K(r)rr′2(2-r2)K(r).
It follows from Lemma2.1 (1) and (2.6) that f′(r)>0 for r∈(0,1), which implies that f(r) is strictly increasing in (0,1).
Therefore, inequality (2.3) follows from (2.4) and (2.5) together with the monotonicity of f(r).Lemma 2.3.
Inequality(2.7)23πrK(r)>log⁡(2+3r2-3r)
holds for all r∈(0,1).Proof.
Let(2.8)g(r)=23πrK(r)-log⁡(2+3r2-3r).
Then simple computations lead to
(2.9)g(0)=0,(2.10)g′(r)=23π(K(r)+rE(r)-r′2K(r)rr′2)-434-3r2=23π(1+3r′2)(1+3r′2r′3/2E(r)r′1/2-2π).
Clearly the functionr→(1+3r2)/r3/2 is strictly decreasing from (0,1) onto (4,+∞). Then (2.10) and Lemma 2.1 (2) lead to the conclusion that g′(r)>0 for r∈(0,1). Thus, g(r) is strictly increasing in (0,1).
Therefore, inequality (2.7) follows from (2.8) and (2.9) together with the monotonicity of g(r).


## 3. Proof of Theorems1.1–1.3

Proof of Theorem1.1.
Letλ=1/4, then from the monotonicity of the function f1(x)=H(xa+(1-x)b,xb+(1-x)a) in [0,1/2] we know that to prove inequality (1.10) we only need to prove that
(3.1)AG(a,b)<H(λa+(1-λ)b,λb+(1-λ)a)
for all a,b>0 with a≠b.
From (1.1) and (1.7) we clearly see that both AG(a,b) and H(λa+(1-λ)b,λb+(1-λ)a) are symmetric and homogeneous of degree 1. Without loss of generality, we can assume that a=1>b. Let t=b∈(0,1) and r=(1-t)/(1+t), then from (1.5) we have
(3.2)H(λa+(1-λ)b,λb+(1-λ)a)-AG(a,b)=(t+3)(3t+1)8(1+t)-π2K(t′).
Let(3.3)F(t)=(t+3)(3t+1)8(1+t)-π2K(t′).
Then making use of (2.2) we get
(3.4)F(t)=(2+r)(2-r)4(1+r)-π2(1+r)K(r)=π8(1+r)K(r)F1(r),
where F1(r)=(2/π)(4-r2)𝒦(r)-4. Note that
(3.5)F1(r)=∑n=0∞(1/2,n)2(n!)2r2n(4-r2)-4=4r2∑n=0∞(1/2,n+1)2[(n+1)!]2r2n-r2∑n=0∞(1/2,n)2(n!)2r2n=∑n=0∞(1/2,n)2[(n+1)!]2(3n2+2n)r2(n+1)>0.
Therefore, inequality (3.1) follows from (3.2)–(3.5).Next, we prove that the parameterp=λ=1/4 is the best possible parameter in (0,1/2) such that inequality (1.10) holds for all a,b>0 with a≠b.Since for0<p<1/2 and small x>0,(3.6)AG(1,1-x)=π2K(2x-x2)=1-12x-116x2+o(x3),(3.7)H(p(1-x)+1-p,(1-p)(1-x)+p)=1-12x+(-p2+p-14)x2+o(x3).It follows from (3.6) and (3.7) that inequality AG(1,1-x)≤H(p(1-x)+1-p,(1-p)(1-x)+p) holds for small x only p≥1/4.Remark 3.1.
For0<p<1/2 and x>0, one has
(3.8)lim⁡x→0H(px+1-p,(1-p)x+p)AG(1,x)=lim⁡x→04[px+1-p][(1-p)x+p](1+x)πK(x′)=+∞.
Equation (3.8) implies that there does not exist p∈(0,1/2) such that AG(1,x)>H(px+1-p,(1-p)x+p) for all x∈(0,1).Proof of Theorem1.2.
Letμ=1/2-2/4, then from the monotonicity of the function f2(x)=G(xa+(1-x)b,xb+(1-x)a) in [0,1/2] we know that to prove inequality (1.11) we only need to prove that
(3.9)AG(a,b)<G(μa+(1-μ)b,μb+(1-μ)a)
for all a,b>0 with a≠b.
From (1.1) and (1.8) we clearly see that both AG(a,b) and G(μa+(1-μ)b,μb+(1-μ)a) are symmetric and homogeneous of degree 1. Without loss of generality, we can assume that a=1>b. Let t=b∈(0,1) and r=(1-t)/(1+t), then from (1.5) we have
(3.10)G(μa+(1-μ)b,μb+(1-μ)a)-AG(a,b)=[μ+(1-μ)t][μt+(1-μ)]-π2K(t′).
Let(3.11)G(t)=[μ+(1-μ)t][μt+(1-μ)]-π2K(t′).
Then making use of (2.2) we have
(3.12)G(t)=π2(1+r)K(r)[2πK(r)1-12r2-1].
Therefore, inequality (3.9) follows from (3.10)–(3.12) together with Lemma 2.2.Next, we prove that the parameterq=μ=1/2-2/4 is the best possible parameter in (0,1/2) such that inequality (1.11) holds for all a,b>0 with a≠b.Since for0<q<1/2 and small x>0,(3.13)G(q(1-x)+1-q,(1-q)(1-x)+q)=1-12x+18(-4q2+4q-1)x2+o(x3).It follows from (3.6) and (3.13) that inequality AG(1,1-x)≤G(q(1-x)+1-q,(1-q)(1-x)+q) holds for small x only q≥1/2-2/4.Remark 3.2.
For0<q<1/2 and x>0, one has
(3.14)lim⁡x→0G(qx+1-q,(1-q)x+q)AG(1,x)=lim⁡x→02π[qx+1-q][(1-q)x+q]K(x′)=+∞.
Equation (3.14) implies that there does not exist q∈(0,1/2) such that AG(1,x)>G(qx+1-q,(1-q)x+q) for all x∈(0,1).Proof of Theorem1.3.
Letβ=1/2-3/4, then from the monotonicity of f3(x)=L(xa+(1-x)b,xb+(1-x)a) in [0,1/2] we know that to prove inequality (1.12) we only need to prove that
(3.15)AG(a,b)<L(βa+(1-β)b,βb+(1-β)a)
for all a,b>0 with a≠b.
From (1.1) and (1.9) we clearly see that both AG(a,b) and L(βa+(1-β)b,βb+(1-β)a) are symmetric and homogeneous of degree 1. Without loss of generality, we can assume that a=1>b. Let t=b∈(0,1) and r=(1-t)/(1+t), then from (1.5) one has
(3.16)L(βa+(1-β)b,βb+(1-β)a)-AG(a,b)=3(1-t)2log⁡[((2-3)t+2+3)/((2+3)t+2-3)]-π2K(t′).
Let(3.17)J(t)=3(1-t)2log⁡[((2-3)t+2+3)/((2+3)t+2-3)]-π2K(t′).
Then from (2.2) we get
(3.18)J(t)=π2(1+r)K(r)log⁡((2+3r)/(2-3r))g(r),
where g(r) is defined as in Lemma 2.3.
Therefore, inequality (3.15) follows from (3.16)–(3.18) together with Lemma 2.3.Next, we prove that the parameters=β=1/2-3/4 is the best possible parameter in (0,1/2) such that inequality (1.12) holds for all a,b>0 with a≠b.Since for0<s<1/2 and small x>0,(3.19)L(s(1-x)+1-s,(1-s)(1-x)+s)=1-12x+112(-4s2+4s-1)x2+o(x3).It follows from (3.6) and (3.19) that inequality AG(1,1-x)≤L(s(1-x)+1-s,(1-s)(1-x)+s) holds for small x only s≥1/2-3/4.Remark 3.3.
For0<s<1/2 and x>0, one has
(3.20)lim⁡x→0L(sx+1-s,(1-s)x+s)AG(1,x)=lim⁡x→02πK(x′)(1-2s)(1-x)log⁡[(sx+1-s)/((1-s)x+s)]=+∞.
Equation (3.20) implies that there exist no values s∈(0,1/2) such that AG(1,x)>L(sx+1-s,(1-s)x+s) for all x∈(0,1).



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