A note on mean inequalities

Introduction

Throughout this note, we consider mean of $\mathit{x}=(x_1,\dots ,x_n)$, 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: $$\mu (\mathit{x};g)=g^{-1}(\frac{1}{n}\sum _{i=1}^{n}g(x_i)),$$ 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 $g(x)=x,\log (x),1/x,\text{and}{x}^2$, respectively.

Next, we introduce the so-called power mean. We define a family of functions $$P(x;\alpha )=x^{\alpha },\text{if}\alpha \ne 0,\text{otherwise}=\log x,$$ where the domain is $x\in (0,+\mathrm{\infty })$. The power mean of $\mathit{x}=(x_1,\dots ,x_n)$ is $$\mu (\mathit{x};P(x;\alpha )).$$

The following facts are well known:

• [F1] $$\underset{\alpha \to -\mathrm{\infty }}{lim}\mu (\mathit{x};P(x;\alpha ))=min{x}_1,\dots ,x_n.$$

• [F2] $$\underset{\alpha \to +\mathrm{\infty }}{lim}\mu (\mathit{x};P(x;\alpha ))=max{x}_1,\dots ,x_n.$$

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

The above fact [F3] implies that $$\mu (\mathit{x};P(x;-1))\le \mu (\mathit{x};P(x;0))\le \mu (\mathit{x};P(x;1))\le \mu (\mathit{x};P(x;2)),$$ that is, $$\text{H-mean}\le \text{G-mean}\le \text{A-mean}\le \text{Q-mean},$$ 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 $\mu (\mathit{x};g)\le \mu (\mathit{x};h),$ where both $g(\cdot )$ and $h(\cdot )$ are continuous monotone functions and their definite integrals on any interval $[c_1,c_2]$ ($0<c_1<c_2<+\mathrm{\infty }$) exist, then for any $0<a<b<+\mathrm{\infty }$, $$g^{-1}(\frac{1}{b-a}{\int }_a^{b}g(x)dx)\le h^{-1}(\frac{1}{b-a}{\int }_a^{b}h(x)dx).$$

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

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

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

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

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

Example 2: Since $$\mu (\mathit{x};P(x;p))\le \mu (\mathit{x};P(x;q)),$$ 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 $0<a<b<+\mathrm{\infty }$, $p<q$, $p\ne -1$ and $q\ne -1$.

References

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