# 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.

## Some Definitions of Means

**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, and harmonic mean of \(a,b,c\) are 8, 5, 3, respectively. What is the value of \( a^2 + b^2 + c^2 \)?

This problem is posed by Matt.

**Details and Assumptions:**

The arithmetic mean, geometric mean, and harmonic mean of \(n\) numbers \(a_1, a_2, \ldots, a_n\) are (respectively)

\[ \frac {\sum_{i=1}^{n} a_n}{n},\quad \sqrt[n]{\prod_{i=1}^{n} a_n},\quad \left( \frac{n} { \sum_{i=1}^{n} \frac{1}{a_n} } \right) . \]

## QM-AM-GM-HM Inequality

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

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

as desired.

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

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

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 \( \sqrt{\frac{a^2+b^2}{2}} \geq \frac{a+b}{2} .\)

Squaring the expression on each side, we have

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

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 \( \frac{a+b}{2} \geq \sqrt{ab} .\)

Subtracting the right side from the left side gives

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

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

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

Subtracting the right side from the left side gives

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

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

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

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

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: \[\]

\[\begin{array} &1, &\frac{a}{a-1}, &\frac{a+1}{a}. \end{array}\]

The differences among the three expressions are

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

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

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

As shown above, the 3 colored squares have the side lengths of \(a<b<c\) while the blue rectangle has the length of \(a+b+c\) and the width of \(d\).

If the blue rectangle's area is equal to the sum of 3 squares' areas combined, which of the following statements is correct?

## Power Mean Inequality

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 \( m > n. \)

Power Mean InequalityFor 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 \( a_1 = \cdots = a_k. \)

**Cite as:**Power Mean Inequality (QAGH).

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/power-mean-qagh/