Cauchy's Mean Theorem

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 $$G_n$$ be the geometric mean of $$x_1, x_2, \ldots, x_n$$.

Then $$A_n \ge G_n$$.

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

$$G_n = \left({\prod_{k=1}^n x_k}\right)^{1/n}$$.

We prove the result by induction:

For all $$n \in \mathbb{N}^*$$, let $$P \left({n}\right)$$ be the proposition "For all positive numbers $$x_1, x_2, \ldots, x_n: A_n \ge G_n$$."


 * $$P(1)$$ is true, as this just says $$\frac {x_1} 1 \ge x_1^{1/1}$$ which is trivially true.

Basis for the Induction

 * $$P(2)$$ is the case $$\frac {x_1 + x_2} 2 \ge \sqrt{x_1 x_2}$$.

As $$x_1, x_2 > 0$$ we can take their square roots and do the following:

$$ $$ $$

This is our basis for the induction.

Induction Hypothesis
Now we show that:


 * if $$P \left({2^k}\right)$$ is true, where $$k \ge 1$$, then it logically follows that $$P \left({2^{k+1}}\right)$$ is true;
 * if $$P \left({k}\right)$$ is true, where $$k \ge 2$$, then it logically follows that $$P \left({k-1}\right)$$ is true.

The result will follow by Backwards Induction.

So this is our first induction hypothesis:

$$A_{2^k} \ge G_{2^k}$$.

Then we need to show:

$$A_{2^{k+1}} \ge G_{2^{k+1}}$$.

Induction Step
This is our induction step:

Let $$m = 2^k$$. Then $$2^{k+1} = 2m$$.

Since $$P \left({m}\right)$$ is true, $$\left({x_1 x_2 \cdots x_m}\right)^{1/m} \le \frac 1 m \left({x_1 + x_2 + \cdots + x_m}\right)$$

Also, $$\left({x_{m+1} x_{m+2} \cdots x_{2m}}\right)^{1/m} \le \frac 1 m \left({x_{m+1} + x_{m+2} + \cdots + x_{2m}}\right)$$

But we have $$P(2)$$, so:

$$\left({\left({x_1 x_2 \cdots x_m}\right)^{1/m} \left({x_{m+1} x_{m+2} \cdots x_{2m}}\right)^{1/m}}\right)^{1/2} \le \frac 1 2 \left({\frac {x_1 + x_2 + \cdots + x_m} m \frac {x_{m+1} + x_{m+2} + \cdots + x_{2m}} m}\right)$$.

So $$\left({x_1 x_2 \cdots x_{2m}}\right)^{1/2m} \le \frac 1 2 \left({\frac {x_1 + x_2 + \cdots + x_{2m}} {2m}}\right)$$.

So $$P \left({2m}\right) = P \left({2^{k+1}}\right)$$ holds.

So $$P \left({2^n}\right)$$ holds for all $$n$$ by induction.


 * Now suppose $$P \left({k}\right)$$ holds. Then:

$$ $$ $$ $$

So $$P \left({k}\right) \Longrightarrow P \left({k-1}\right)$$ and the result follows by Backwards Induction.

Therefore $$A_n \ge G_n$$ for all $$n$$.