Exponential Inequalities for Means

Dorin Marghidanu,
February 4, 2016

We prove elegant inequalities between the exponents of the arithmetic $\displaystyle A_n=\frac{1}{n}\sum_{i=1}^{n}a_n,\;$ geometric $\displaystyle G_n=\sqrt[n]{\prod_{i=1}^{n}a_n},\;$ and the harmonic $\displaystyle H_n=\frac{n}{\sum_{i=1}^{n}\displaystyle \frac{1}{a_n}}\;$ means:

Exponential inequalities for means

Proof

Consider the function $f:\;(0,\infty\rightarrow\mathbb{R},\;$ $f(x)=x^{\displaystyle\frac{1}{x}}.\;$ Since $f'(x)=x^{\displaystyle\frac{1}{x}-2}(1-\ln x),\;$ for $x\gt e,\;$ we have $f'(x)\lt 0.\;$ Hence, the function $f\;$ is strictly decreasing on $(e,\infty).\;$ In addition,

$\min\{a_i\}_{i=1}^{n}\le A_n\le G_n\le H_n\le\max\{a_i\}_{i=1}^{n}.$

It follows that, for $a_1, a_2, \ldots, a_n\ge e,\;$ $f(A_n)\le f(G_n)\le f(H_n),\;$ i.e.,

$A_n^{\displaystyle\frac{1}{A_n}}\le G_n^{\displaystyle\frac{1}{G_n}}\le H_n^{\displaystyle\frac{1}{H_n}}.$

By raising the inequalities to the power of $A_nG_nH_n\;$ we obtain the desired inequalities. The case of $a_1, a_2, \ldots, a_n\le e\;$ is similar.

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