A note on mean inequalities

Introduction

Throughout this note, we consider mean of ${\boldsymbol x}=(x_1, \ldots, x_n)$, where $n$ is an arbitrary positive integer and real number $x_i>0$ for $1\le i \le n$. Carvalho [1] discusses Kolmogorov’s general definition of mean: \begin{equation} \mu({\boldsymbol x}; g)= g^{-1}\left(\frac{1}{n}\sum_{i=1}^n g(x_i)\right), \end{equation} where $g(\cdot)$ is a continuous monotone function and $g^{-1}(\cdot)$ is its inverse. Arithmetic, geometric, harmonic, and quadratic means are special cases of the above, because we can let $g(x)=x,\ \log(x),\ 1/x,\ \hbox{and}\ x^2$, respectively.

Next, we introduce the so-called power mean. We define a family of functions $$ P(x; \alpha)= x^\alpha,\ \hbox{if}\ \alpha \neq 0,\ \hbox{otherwise}
=\log x, $$ where the domain is $x \in (0, +\infty)$. The power mean of ${\boldsymbol x}=(x_1, \ldots, x_n)$ is $$ \mu\left({\boldsymbol x}; P(x; \alpha)\right). $$

The following facts are well known:

• [F1] $$\lim_{\alpha \rightarrow -\infty}\mu\left({\boldsymbol x}; P(x; \alpha)\right)=\min{x_1, \ldots, x_n}.$$

• [F2] $$\lim_{\alpha \rightarrow +\infty}\mu\left({\boldsymbol x}; P(x; \alpha)\right)=\max{x_1, \ldots, x_n}.$$

• [F3] As a function of $\alpha$, $\mu\left({\boldsymbol x}; P(x; \alpha)\right)$ is increasing when $\alpha$ increases.

The above fact [F3] implies that \begin{equation} \mu\left({\boldsymbol x}; P(x; -1)\right)\le \mu\left({\boldsymbol x}; P(x; 0)\right)\le \mu\left({\boldsymbol x}; P(x; 1)\right)\le \mu\left({\boldsymbol x}; P(x; 2)\right), \end{equation} that is, $$ \hbox{H-mean}\le \hbox{G-mean}\le \hbox{A-mean}\le \hbox{Q-mean}, $$ where ‘H’, ‘G’, ‘A’ and ‘Q’ stand for ‘harmonic’, ‘geometric’, ‘arithmatic’ and ‘quadratic’, respectively.

In this note, based on the above, an inequality is derived in Theorem 1, which, I believe, deserves to be widely known. Two examples are shown as special cases of Theorem 1.

Main results

Theorem 1: If $\mu({\boldsymbol x}; g)\le \mu({\boldsymbol x}; h),$ where both $g(\cdot)$ and $h(\cdot)$ are continuous monotone functions and their definite integrals on any interval $[c_1, c_2]$ ($0<c_1<c_2<+\infty$) exist, then for any $0<a<b<+\infty$, \begin{equation} g^{-1}\left(\frac{1}{b-a}\int_a^b g(x)dx\right)\le h^{-1}\left(\frac{1}{b-a}\int_a^b h(x)dx\right). \end{equation}

Proof: Take an arbitrary positive integer $n$, and let $$ x_i=a+i(b-a)/n,\ \hbox{for}\ i=1, \ldots, n. $$ Let $$ {\boldsymbol x}_n=(x_1, \ldots, x_n). $$ According to the assumption,

$$ \mu({\boldsymbol x}_n; g)\le \mu({\boldsymbol x}_n; h), $$

thus taking limit on the both sides leads to the desired result.

Example 1: Using Theorem 1 and noticing $$ \mu\left({\boldsymbol x}; P(x; -2)\right)\le \mu\left({\boldsymbol x}; P(x; -1)\right) $$ together with the result $\hbox{H-mean}\le \hbox{G-mean} \le \hbox{A-mean} \le \hbox{Q-mean}$, we have the following inequality chain. For any $0<a<b<+\infty$, $$ \sqrt{ab}\le \frac{b-a}{\log b- \log a}\le \exp\left(\frac{b\log b - a\log a}{b-a}-1\right) $$

$$ \le\frac{b+a}{2}\le \sqrt{\frac{b^2+ab+a^2}{3}}. $$

Example 2: Since $$ \mu\left({\boldsymbol x}; P(x; p)\right)\le \mu\left({\boldsymbol x}; P(x; q)\right), $$ if $p<q$, by Theorem 1, we derive $$ \sqrt[p]{\frac{b^{p+1}-a^{p+1}}{(b-a)(p+1)}}\le \sqrt[q]{\frac{b^{q+1}-a^{q+1}}{(b-a)(q+1)}}, $$ where $0<a<b<+\infty$, $p<q$, $p\neq -1$ and $q\neq -1$.

References

[1] Miguel de Carvalho, Mean, What do You Mean? The American Statistician 70 (2016) 270–274.