Ky Fan inequalities for bivariate means

TL;DR

This paper establishes sufficient conditions under which certain bivariate means satisfy Ky Fan inequalities, contributing to the theoretical understanding of inequalities involving symmetric, homogeneous means.

Contribution

It provides new theoretical conditions ensuring Ky Fan inequalities hold for classes of bivariate, homogeneous, symmetric means.

Findings

Derived sufficient conditions for Ky Fan inequalities to hold.

Applied inequalities to specific classes of means.

Enhanced understanding of mean inequalities in mathematical analysis.

Abstract

In this note we give sufficient conditions for bivariate, homogeneous, symmetric means $M$ and $N$ to satisfy Ky Fan inequalities $$\frac{M}{M'}<\frac{N}{N'}\qquad\text{ and }\quad \frac{1}{M}-\frac{1}{M'}<\frac{1}{N}-\frac{1}{N'}.$$

Equations

$$\frac{M}{M^{\prime }}<\frac{N}{N^{\prime }}\text{ and }\frac{1}{M}-\frac{1}{M^{\prime }}<\frac{1}{N}-\frac{1}{N^{\prime }}.$$

$$\mathsf{G}(x_1,\dots ,x_n)=\prod _{i=1}^n{x}_i^{1/n},\mathsf{A}(x_1,\dots ,x_n)=\frac{1}{n}\sum _{i=1}^n{x}_i.$$

$$\frac{\mathsf{G}(x_1,\dots ,x_n)}{\mathsf{G}(1-x_1,\dots ,1-x_n)}\le \frac{\mathsf{A}(x_1,\dots ,x_n)}{\mathsf{A}(1-x_1,\dots ,1-x_n)}.$$

$$\frac{\mathsf{G}}{{\mathsf{G}}^{\prime }}\le \frac{\mathsf{L}}{{\mathsf{L}}^{\prime }}\le \frac{\mathsf{P}}{{\mathsf{P}}^{\prime }}\le \frac{\mathsf{A}}{{\mathsf{A}}^{\prime }}\le \frac{\mathsf{NS}}{\mathsf{N}{\mathsf{S}}^{\prime }}\le \frac{\mathsf{T}}{{\mathsf{T}}^{\prime }},$$

$$\frac{M(x,y)}{M(1-x,1-y)}\le \frac{N(x,y)}{N(1-x,1-y)}$$

$$\min (x,y)\le M(x,y)\le \max (x,y).$$

$$M(\lambda x,\lambda y)=\lambda M(x,y).$$

$$\frac{z}{1+z}\le m(z)\le \frac{z}{1-z}$$

$$M(x,y)=\frac{|x-y|}{2m\left(\frac{|x-y|}{x+y}\right)}$$

$$\frac{M(x,y)}{M(1-x,1-y)}\le \frac{N(x,y)}{N(1-x,1-y)}$$

$$\frac{M(x,y)}{N(x,y)}=\frac{n}{m}\left(\frac{|x-y|}{x+y}\right)<\frac{n}{m}\left(\frac{|x^{\prime }-y^{\prime }|}{x^{\prime }+y^{\prime }}\right)=\frac{M(x^{\prime },y^{\prime })}{N(x^{\prime },y^{\prime })},$$

$$\frac{M(x,y)}{M(1-x,1-y)}\le \frac{N(x,y)}{N(1-x,1-y)}$$

$$m(z)=\frac{z}{M(1+z,1-z)}.$$

$$\frac{n}{m}(z)=\frac{M(1+z,1-z)}{N(1+z,1-z)}=\frac{M\left(1,\frac{1-z}{1+z}\right)}{N\left(1,\frac{1-z}{1+z}\right)}=\frac{M(1,s)}{N(1,s)}=q(s).$$

$${\mathsf{A}}_r(x,y)=\{\begin{array}{ll}{\left(\frac{x^r+y^r}{2}\right)}^{1/r}& r\ne 0,\\ \sqrt{xy}& r=0\end{array}$$

$$\frac{{\mathsf{A}}_r}{{\mathsf{A}}_r^{\prime }}\le \frac{{\mathsf{A}}_s}{{\mathsf{A}}_s^{\prime }}.$$

$$\mathrm{sgn}\frac{d}{dt}\log \frac{{\mathsf{A}}_r(1,t)}{{\mathsf{A}}_s(1,t)}=\mathrm{sgn}\frac{t^r-t^s}{t(t^r+1)(t^s+1)}=\mathrm{sgn}(s-r),$$

$$\mathrm{sgn}\frac{d}{dt}\log \frac{{\mathsf{A}}_r(1,t)}{{\mathsf{A}}_0(1,t)}=\mathrm{sgn}\frac{t^r-1}{t(2t^r+1)}=-\mathrm{sgn}r,$$

$$\mathsf{T}(x,y)=\frac{|x-y|}{2\arctan \frac{|x-y|}{x+y}}.$$

$$\frac{d}{dz}\frac{\mathsf{t}}{\mathsf{q}}(z)=\frac{z-\arctan z}{z^2\sqrt{1+z^2}}>0,$$

$$\frac{\mathsf{T}}{{\mathsf{T}}^{\prime }}<\frac{\mathsf{Q}}{{\mathsf{Q}}^{\prime }}.$$

$$\mathsf{He}(x,y)=\frac{x+\sqrt{xy}+y}{3}.$$

$$q(t)=\frac{\mathsf{He}(1,t)}{{\mathsf{A}}_{2/3}(1,t)}=\frac{2\sqrt{2}}{3}\frac{t+\sqrt{t}+1}{(t^{2/3}+1{)}^{3/2}}$$

$$\frac{dq}{dt}(t)=\frac{\sqrt{2}}{3}\frac{(1-t^{1/6}{)}^3(1+t^{1/6})}{(1+t^{2/3}{)}^{5/2}{t}^{1/2}}>0.$$

$$\frac{\mathsf{He}(1,t)}{{\mathsf{A}}_{1/2}(1,t)}$$

$$=\frac{4}{3}\left[1-\frac{1}{\sqrt{t}+2+\frac{1}{\sqrt{t}}}\right]$$

$$\frac{{\mathsf{A}}_{1/2}}{{\mathsf{A}}_{1/2}^{\prime }}<\frac{\mathsf{He}}{{\mathsf{He}}^{\prime }}<\frac{{\mathsf{A}}_{2/3}}{{\mathsf{A}}_{2/3}^{\prime }}.$$

$$q(t)=\frac{{\mathsf{A}}_{1/3}(1,t)}{\mathsf{L}(1,t)}=\frac{(t^{1/3}+1{)}^3}{8(t-1)}\log t.$$

$$\frac{dq}{dt}(t)=\frac{(t^{1/3}+1{)}^2[t^{4/3}+t-t^{1/3}-1-(t+t^{1/3})\log t]}{8t(t-1{)}^2}.$$

$$t^{4/3}+t$$

$$=-\sum _{n=5}^{\infty }\frac{n^2-5n+12}{n(n-1)(n-2)(n-3)}(1-s{)}^n<0,$$

Full text

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1. Introduction

Denote by $\mathsf{G}$ and $\mathsf{A}$ the geometric and arithmetic means of nonnegative numbers $x_{i},\ i=1,\dots,n$

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The classical Ky Fan inequality states that if $0<x_{i}\leq 1/2$, then

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This result has been published in 1961 and generalized in many directions. For more information see [1].

This note is inspired by the result of Neuman and Sándor [2] where we find the following sequence of inequalities

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(here ${\mathsf{L}},{\mathsf{P}},{\mathsf{NS}},{\mathsf{T}}$ stand for the logarithmic, first Seiffert, Neuman-Sándor and second Seiffert means of arguments $0<x,y\leq 1/2$, and prime denotes the same mean with arguments $1-x$ and $1-y$).

The aim of this note is to establish the conditions for two symmetric, homogeneous means $M,N$ under which the Ky Fan inequality

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holds.

2. Definition and notation

A function $M:\mathbb{R}_{+}^{2}\to\mathbb{R}$ is called a mean if for all $x,y>0$ holds

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Mean is symmetric if for all $M(x,y)=M(y,x)$ for all $x,y>0$ and homogeneous if for all $\lambda>0$

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A function $m:(0,1)\to\mathbb{R}$ satisfying

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is called a Seiffert function.

Let us recall a result from [7]

Theorem 2.1**.**

The formula

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establishes a one-to-one correspondence between the set of symmetric, homogeneous means and the set of Seiffert functions.

If there is no risk of ambiguity we shall skip the argument of means. For $0<x,y\leq 1/2$ and an arbitrary mean $M$ we shall denote $M(1-x,1-y)$ by $M^{\prime}$.

3. Ky Fan inequalities

Our first theorem shows that the Ky Fan inequality is a consequence of monotonicity of the ratio of Seiffert means.

Theorem 3.1**.**

Let $M,N$ be two symmetric, homogeneous means and $m,n$ be their Seiffert functions. If the function $\frac{n}{m}$ is decreasing, then for all $0<x,y\leq 1/2$ the Ky Fan inequality

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holds.

Proof.

Denote $x^{\prime}=1-x,y^{\prime}=1-y$. Clearly $|x-y|=|x^{\prime}-y^{\prime}|$ and $x+y\leq x^{\prime}+y^{\prime}$, thus $\frac{|x-y|}{x+y}\geq\frac{|x^{\prime}-y^{\prime}|}{x^{\prime}+y^{\prime}}$. By Theorem 2.1 we obtain

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which concludes the proof. ∎

Sometimes it is more convenient to investigate the ratio of means instead of their Seiffert functions.

Corollary 3.2**.**

In the assumptions of Theorem 3.1 if the function $q(t)=\frac{M(1,t)}{N(1,t)}$ increases in $(0,1)$, then for all $0<x,y\leq 1/2$ the Ky Fan inequality

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holds.

Proof.

Let us recall the formula connecting the mean and its Seiffert function ([7])

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This gives:

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Here $s=\frac{1-z}{1+z}$ decreases from $1$ to [math] as $z$ travels in the opposite direction, so $n/m$ decreases if and only if $q$ increases. ∎

Let us illustrate our results with some examples:

Example 3.1*.*

This result was proved by Chan, Goldberg and Gonek in [3]: if

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is a power mean of order $r$, then for $r<s$ holds

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Indeed, for $0<t<1$ and $rs\neq 0$

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else if $s=0$, then

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so the inequality (3) is true by Corollary 3.2.

Example 3.2*.*

The last term in the chain of inequalities (1) contains the second Seiffert mean

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Let $\mathsf{Q}(x,y)=\sqrt{\frac{x^{2}+y^{2}}{2}}$ be the quadratic (or root mean square).

We have $\mathsf{t}(z)=\arctan z$ and $\mathsf{q}(z)=\frac{z}{\sqrt{1+z^{2}}}$ and

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so by Theorem 3.1 we can extend the chain (1)

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Example 3.3*.*

Consider now the Heronian mean

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For $0<t<1$ we have

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and

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We also see that the quotient

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decreases in $(0,1)$. Therefore by Corollary 3.2 we have

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Example 3.4*.*

We know ([5]) that the logarithmic mean $\mathsf{L}(x,y)=\frac{x-y}{\log x-\log y}$ satisfies the inequality $\mathsf{L}<\mathsf{A}_{1/3}$. Consider

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Its derivative equals

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To evaluate the sign of the expression in square brackets substitute $t=s^{3}$ and calculate its Taylor series at $s=1$

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which proves the inequality

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Example 3.5*.*

There is another mean similar to the Seiffert means that lies between the arithmetic and the first Seiffert mean:

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Their respective Seiffert functions are $\arcsin,\sinh$ and $\mathrm{id}$. The hyperbolic sine is convex for positive arguments, therefore $\frac{\sinh z}{z}$ increases as a divided difference of a convex function. So by Theorem 3.1 we conclude

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Example 3.6*.*

It is known [4] that $\mathsf{A}_{1/2}<\mathsf{P}$. We have $\mathsf{a}_{1/2}(z)=\frac{2z}{1+\sqrt{1-z^{2}}}$. To check the monotonicity of $\mathsf{p}/\mathsf{a}_{1/2}$ we substitute $z=\sin t$ to get $\frac{t}{2\sin t}(1+\cos t)$. Its derivative $\frac{t-\sin t}{2\cos t-2}$ is negative, so

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Example 3.7*.*

The tangent is also a Seiffert function. It satisfies the inequality $\tan z<\operatorname{artanh}z$ for $0<z<1$, so their means satisfy

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Let us investigate the quotient of the Seiffert functions:

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and using Taylor expansion we obtain

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because all the coefficients in square brackets are nonnegative. Thus by Theorem 3.1

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4. Harmonic Ky Fan inequalities

In [6] the autors derive the inequalities of type

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from the inequalities $\mathsf{P}/\mathsf{P}^{\prime}<\mathsf{A}/\mathsf{A}^{\prime}$ using some elementary algebraic transformation. Here we show an approach that bases on the concept of Seiffert functions.

Suppose $M$ ans $N$ are two symmetric and homogeneous means. Using Theorem 2.1 the inequality

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can be written as

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Since $2-x-y>x+y$ for $0<x,y<1/2$ we can formulate the following result.

Theorem 4.1**.**

Suppose the means $M$ and $N$ are symmetric and homogeneous and $m$ and $n$ are their Seiffert means. If the function $m-n$ decreases, then the Ky Fan inequality

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holds.

Taking into account the formula (2) and setting $s=(1+z)/(1-z)$ we get the following

Corollary 4.2**.**

In the assumptions of Theorem 4.1 if the function

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descreases for $s>1$, then the inequality

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holds.

Example 4.1*.*

Consider the chain of inequalities between Seiffert functions (see [7, Lemma 3.1]) valid for $0<z<1$

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The mean corresponding to the second function is called Neuman-Sándor mean — $\mathsf{NS}$ and the fourth one generates the second Seiffert mean — $\mathsf{T}$.

Since $\cosh^{2}z-1=(\cosh z+1)(\cosh z-1)>2\cdot\frac{z^{2}}{2}$

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Inequality $\cos z>1-z^{2}/2$ leads to

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On the other hand $\cos z<1-z^{2}/2+z^{4}/6$ , so

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And finally

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Therefore by Theorem 4.1

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Example 4.2*.*

On the other side of the arithmetic mean there are two chains of inequalities for Seiffert means involving sine and tangent (see [7, Lemma 3.2]):

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The two Seiffert functions in curly brackets are not comparable, arcsine defines the first Seiffert mean and inverse hyperbolic tangent corresponds to the logarithmic mean.

The difference $\sinh z-z$ increases, because this is the gap between a convex function and its supporting line.

To show that $\frac{d}{dz}(\tan z-\sinh z)=\frac{1-\cos^{2}z\cosh z}{\cos^{2}z}$ is positive note that

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The function $f$ vanishes at $z=0$ and

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because $\tan z>z>\tanh z$. Thus $f^{\prime}$ is positive, and so is $f$.

The difference between inverse hyperbolic tangent and tangent also increases, since

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To deal with the lower chain of inequalities (4) note that

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Now

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because using (5)

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Comparison of the last pair is simple

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Now we can use Theorem 4.1 to write the chain of inequalities

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NOTE: In the proof of Theorems 3.1 we use monotonicity of $\frac{m}{n}$ to obtain inequality between $\frac{m}{n}\left(\frac{y-x}{x+y}\right)$ and $\frac{m}{n}\left(\frac{y-x}{2-x-y}\right)$. It is natural to ask whether the inverse is true, i.e. if for all $0<x<y<1/2$ the inequality

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holds then $f$ needs to be monotone. The counterexample has been produced by tometomek91 - user of the mathematical portal matematyka.pl.

It is clear that $\frac{y-x}{x+y}>\frac{y-x}{2-x-y}$. Moreover $\sup_{0<x<y<1/2}\frac{y-x}{2-x-y}=\frac{1}{3}$. Therefore every function $f$ satisfying the two conditions:

• •

$f$ is increasing on $(0,1/3)$

• •

$f(x)\geq\lim_{t\nearrow{1/3}}f(t)$ for $x\in[1/3,1]$

satisfies (6).

5. Acknowledgement

The author wishes to thank Janusz Matkowski and Monika Nowicka for improvement suggestions and efficient bug and typo hunting.

6. References