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
The arithmetic mean, $A_n$, of $x_1, x_2, \ldots, x_n$ is defined by:
 * $\displaystyle A_n = \frac 1 n \left({\sum_{k \mathop = 1}^n x_k}\right)$

The harmonic mean, $H_n$, of $x_1, x_2, \ldots, x_n$ is defined by:
 * $\displaystyle \frac 1 H_n = \frac 1 n \left({\sum_{k \mathop = 1}^n \frac 1 {x_k}}\right)$

As $\forall k \in \left[{1 \,.\,.\, n}\right]: x_k > 0$, we can express each $x_k$ as a square:
 * $\forall k \in \left[{1 \,.\,.\, n}\right]: x_k = y_k^2$

without affecting the result.

Thus we have:


 * $\displaystyle A_n = \frac 1 n\left({\sum_{k \mathop = 1}^n y_k^2}\right)$


 * $\displaystyle \frac 1 {H_n} = \frac 1 n \left({\sum_{k \mathop = 1}^n \frac 1 {y_k^2}}\right)$

Multiplying $A_n$ by $\displaystyle \frac 1 {H_n}$,

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

The result follows by multiplying both sides by $H_n$ (Real Number Axioms: $\R O2$: compatible with multiplication).

Also see

 * Cauchy's Mean Theorem