Power Mean Inequality (QAGH)

The QM-AM-GM-HM or QAGH inequality generalizes the basic result of the arithmetic mean-geometric mean (AM-GM) inequality, which compares the arithmetic mean (AM) and geometric mean (GM), to include a comparison of the quadratic mean (QM) and harmonic mean (HM), where $f_{\text{QM}}$ denotes the quadratic mean, $f_{\text{AM}}$ denotes the arithmetic mean, $f_{\text{GM}}$ denotes the geometric mean, and $f_{\text{HM}}$ denotes the harmonic mean: $f_{\text{QM}} \geq f_{\text{AM}} \geq f_{\text{GM}} \geq f_{\text{HM}}.$

Furthermore, the power mean inequality extends the QM-AM result to compare higher power means and moments.

Comparisons among various means appear frequently in advanced inequality problems. In addition to the AM-GM inequality, the QM-AM-GM-HM and power mean inequalities are important pieces of the inequality problem solving toolkit.

Arithmetic Mean. Given a list of $k$ positive numbers $a_1, \ldots, a_k,$ one may be interested in a variety of different types of means. By taking the sum divided by the number of values, the arithmetic mean gives an idea of the mean or "typical" value based on a linear weighting. One might imagine that $k$ times the mean gives the sum

$$k f_{\text{AM}} = a_1 + \cdots + a_k,$$

which leads to the definition

$$f_{\text{AM}} = \frac{a_1 + \cdots + a_k}{k}.$$

Quadratic Mean. One can also apply a quadratic weighting. For a different set of values, one might consider $k$ times the square of the mean gives the sum of the squares:

$$k f^2_{\text{QM}} = a_1^2 + \cdots + a_k^2.$$

Thus, finding the sum of the squares divided by the number of the values and then taking the square root gives the quadratic mean or root mean square (RMS):

$$f_{\text{QM}} = \sqrt{\frac{a_1^2 + \cdots + a_k^2}{k}}.$$

Geometric Mean. Alternatively, one might consider the mean with regard to multiplication, with the $k^\text{th}$ power of the mean value equal to the product of the values:

$$f_{\text{GM}}^k = a_1 \cdots a_k.$$

This might lead one to find the product of the values and then take the $k^\text{th}$ root, which yields the geometric mean

$$f_{\text{GM}} = \sqrt[k]{a_1 \cdots a_k}.$$

Harmonic Mean. Finally, one could surmise that $k$ times the reciprocal of the mean might equal the sum of the reciprocals of the values:

$$\frac{k}{f_{\text{HM}}} = \frac{1}{a_1} + \cdots + \frac{1}{a_k}.$$

This leads to the harmonic mean defined as

$$f_{\text{HM}} = \frac{k}{\frac{1}{a_1} + \cdots + \frac{1}{a_k}}.$$

The arithmetic mean-geometric mean (AM-GM) inequality asserts that the the arithmetic mean is never smaller than the geometric mean: $f_{\text{AM}} \geq f_{\text{GM}}.$ It can be used as a starting point to prove the QM-AM-GM-HM inequality.

QM-AM-GM-HM inequality. Given a list of $k$ positive real numbers $a_1, \ldots, a_k$, let $f_{\text{QM}}$ denote the quadratic mean, $f_{\text{AM}}$ denote the arithmetic mean, $f_{\text{GM}}$ denote the geometric mean, and $f_{\text{HM}}$ denote the harmonic mean. Then

$$f_{\text{QM}} \geq f_{\text{AM}} \geq f_{\text{GM}} \geq f_{\text{HM}}.$$

Furthermore, equality is achieved if and only if $a_1 = \cdots = a_k.$

Starting with the AM-GM inequality $f_{\text{AM}} \geq f_{\text{GM}},$ it remains to be proven that $f_{\text{QM}} \geq f_{\text{AM}}$ and $f_{\text{GM}} \geq f_{\text{HM}}.$ The latter directly follows from AM-GM with $\frac1{a_1}, \ldots, \frac1{a_k}$:

$$\frac{\frac{1}{a_1} + \cdots + \frac{1}{a_k}}{k} \geq \sqrt[k]{\frac{1}{a_1 \cdots a_k}}.$$

Taking the reciprocal of both sides yields

$$\sqrt[k]{a_1 \cdots a_k} \geq \frac{k}{\frac{1}{a_1} + \cdots + \frac{1}{a_k}}$$

as desired.

To show the former, one can use the Cauchy-Schwarz inequality to write

$$(a_1 + \cdots + a_k)^2 \leq \big(a_1^2 + \cdots + a_k^2\big)\underbrace{\big(1^2 + \cdots + 1^2\big)}_{k{\text{ times}}}.$$

Dividing both sides by $k^2$ gives

$$\left(\frac{a_1 + \cdots + a_k}{k}\right)^2 \leq \frac{a_1^2 + \cdots + a_k^2}{k},$$

and therefore

$$\frac{a_1 + \cdots + a_k}{k} \leq \sqrt{\frac{a_1^2 + \cdots + a_k^2}{k}}.\ _\square$$

The proof of the condition of equality is left as an exercise.

QM-AM-GM-HM for two variables:

For $a,b > 0,$ it holds that

$$\sqrt{\dfrac{a^2+b^2}{2}} \geq \dfrac{a+b}{2}\geq \sqrt{ab} \geq \dfrac{2ab}{a+b}.$$

When $a$ and $b$ are both positive numbers, prove the inequality $\sqrt{\frac{a^2+b^2}{2}} \geq \frac{a+b}{2} .$

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Squaring the expression on each side, we have

$$\begin{aligned} \left ( \sqrt{\frac{a^2+b^2}{2}} \right )^2 &= \frac{a^2+b^2}{2} \\ \left ( \frac{a+b}{2} \right )^2 &= \frac{a^2+2ab+b^2}{4}. \end{aligned}$$

Subtracting the second equation from the first gives

\[ \begin{align} \frac{a^2+b^2}{2} - \frac{a^2+2ab+b^2}{4} &= \frac{ 2a^2+2b^2-a^2-2ab-b^2}{4} \\\\ &= \frac{a^2-2ab+b^2}{4} \\\\ &= \frac{(a-b)^2}{4} \geq 0.
\end{align} \]

Therefore, $\sqrt{\frac{a^2+b^2}{2}} \geq \frac{a+b}{2}.$ $_\square$

When $a$ and $b$ are both positive numbers, prove the inequality $\frac{a+b}{2} \geq \sqrt{ab} .$

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Subtracting the right side from the left side gives

$$\begin{aligned} \frac{a+b}{2} - \sqrt{ab} &= \frac{a+b-2\sqrt{ab}}{2} \\ &= \frac{\big(\sqrt{a}\big)^2-2\sqrt{a}\sqrt{b}+\big(\sqrt{b}\big)^2}{2} \\ & = \frac{\big(\sqrt{a}-\sqrt{b}\big)^2}{2} \geq 0. \end{aligned}$$

Therefore, $\frac{a+b}{2} \geq \sqrt{ab}.$ $_\square$

When $a$ and $b$ are both positive numbers, prove the inequality $\sqrt{ab} \geq \frac{2ab}{a+b}.$

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Subtracting the right side from the left side gives

$$\begin{aligned} \sqrt{ab}-\frac{2ab}{a+b} &= \frac{\sqrt{ab}(a+b)-2ab}{a+b} \\ &= \frac{\sqrt{ab}\left(a+b-2\sqrt{ab}\right)}{a+b} \\ &= \frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)^2}{a+b} \geq 0. \end{aligned}$$

Therefore, $\sqrt{ab} \geq \frac{2ab}{a+b}.$ $_\square$

When $x$ and $y$ are both positive numbers, what is the minimum value of $(2x+3y)\left(\frac{8}{x} + \frac{3}{y}\right) ?$

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$$\begin{aligned} (2x+3y)\left(\frac{8}{x} + \frac{3}{y}\right) &= 2x \cdot \frac{8}{x} + 2x \cdot \frac{3}{y} + 3y \cdot \frac{8}{x} + 3y \cdot \frac{3}{y} \\ &= \frac{6x}{y} + \frac{24y}{x} + 25 \\ &\geq 2\sqrt{\frac{6x}{y} \times \frac{24y}{x}}+ 25 \\ &= 24+25=49. \end{aligned}$$

Thus, the minimum value of $(2x+3y)\left(\frac{8}{x} + \frac{3}{y}\right)$ is $49.$ $_\square$

When $a>1,$ arrange the following three expressions by magnitude: $$$$

$$\begin{array}{c}&1, &\frac{a}{a-1}, &\frac{a+1}{a}. \end{array}$$

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The differences among the three expressions are

$$\begin{aligned} \frac{a}{a-1}-1 = \frac{a-a+1}{a-1} = \frac{1}{(a-1)} >0 &\Rightarrow \frac{a}{a-1} >1 &\qquad (1) \\ \frac{a+1}{a} - \frac{a}{a-1} = \frac{a^2-1-a^2}{a(a-1)} = -\frac{1}{a(a-1)}<0 &\Rightarrow \frac{a}{a-1} > \frac{a+1}{a} &\qquad (2) \\ \frac{a+1}{a}-1 = \frac{a+1-a}{a} = \frac{1}{a} >0 &\Rightarrow \frac{a+1}{a} > 1. &\qquad (3) \end{aligned}$$

From $(1), (2),$ and $(3),$ the order relation of the three expressions is

$$\frac{a}{a-1} > \frac{a+1}{a} > 1.\ _\square$$

To generalize the arithmetic and quadratic means, one can simply consider any higher power $n$--that is, given an $n^\text{th}$ power weighting, the $n^\text{th}$ power of the mean might be taken to be $k$ times that of the sum of the $n^\text{th}$ powers of some $k$ values:

$$k f^n_n = a_1^n + \cdots + a_k^n.$$

Taking the sum of the $n^\text{th}$ powers divided by the number of the values and then taking the $n^\text{th}$ root gives the $n^\text{th}$ power mean $f_n$:

$$f_n = \sqrt[n]{\frac{a_1^n + \cdots a_k^n}{k}}.$$

The power mean inequality asserts that $f_m \geq f_n$ for $m > n.$

By convention, $f_0 = \sqrt[k]{a_1 a_2 a_3 \cdots a_k}$.

Power Mean Inequality

For positive $a_1, \ldots, a_k,$ with $f_n$ defined as

$$f_n = \sqrt[n]{\frac{a_1^n + \cdots a_k^n}{k}},$$

if $m > n,$ it follows that

$$f_m \geq f_n,$$

with equality holding if and only if $a_1 = \cdots = a_k.$