# Arithmetic Mean is Never Less than Harmonic Mean

(Redirected from AM-HM Inequality)

## Theorem

Let $x_1, x_2, \ldots, x_n \in \R_{> 0}$ be strictly positive real numbers.

Let $A_n$ be the arithmetic mean of $x_1, x_2, \ldots, x_n$.

Let $H_n$ be the harmonic mean of $x_1, x_2, \ldots, x_n$.

Then $A_n \ge H_n$.

## Proof

$A_n$ is defined as:

$\displaystyle A_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n x_k}$

$H_n$ is defined as:

$\displaystyle \frac 1 H_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {x_k} }$

We have that:

$\forall k \in \closedint 1 n: x_k > 0$

From Positive Real has Real Square Root, we can express each $x_k$ as a square:

$\forall k \in \closedint 1 n: x_k = y_k^2$

without affecting the result.

Thus we have:

$\displaystyle A_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n y_k^2}$
$\displaystyle \frac 1 {H_n} = \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {y_k^2} }$

Multiplying $A_n$ by $\dfrac 1 {H_n}$:

 $\displaystyle \frac {A_n} {H_n}$ $=$ $\displaystyle \frac 1 n \paren {\sum_{k \mathop = 1}^n y_k^2} \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {y_k^2} }$ $\displaystyle$ $\ge$ $\displaystyle \frac 1 {n^2} \paren {\sum_{k \mathop = 1}^n \frac {y_k} {y_k} }^2$ Cauchy's Inequality $\displaystyle$ $=$ $\displaystyle \frac 1 {n^2} \paren {\sum_{k \mathop = 1}^n 1}^2$ $\displaystyle$ $=$ $\displaystyle \frac {n^2} {n^2} = 1$

So:

$\dfrac {A_n} {H_n} \ge 1$

and so from Real Number Axioms: $\R \text O 2$: compatible with multiplication:

$A_n \ge H_n$

$\blacksquare$