AM-HM Inequality

Theorem
Let $$x_1, x_2, \ldots, x_n \in \mathbb{R}$$ be real numbers which are all positive.

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 of $$x_1, x_2, \ldots, x_n$$ is defined as:

$$A_n = \frac 1 n \left({\sum_{k=1}^n x_k}\right)$$.

The harmonic mean of $$x_1, x_2, \ldots, x_n$$ is defined as:

$$\frac 1 H_n = \frac 1 n \left({\sum_{k=1}^n \frac 1 {x_k}}\right)$$.

As $$\forall k \in \left[{1 \,. \, . \, n}\right]: x_k > 0$$, we can express all the $$x_k$$'s as squares:

$$\forall k \in \left[{1 \,. \, . \, n}\right]: x_k = y_k^2$$

without affecting the result.

Thus we have:


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


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

Now let's see what happens when we multiply them together:

$$ $$ $$ $$

So $$\frac {A_n} {H_n} \ge 1$$ and the result follows.