Cauchy's Mean Theorem

Theorem
Let $x_1, x_2, \ldots, x_n \in \R$ 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:


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

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


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

We prove the result by induction:

For all $n \in \N^*$, 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:
 * $\dfrac {x_1} 1 \ge x_1^{1/1}$

which is trivially true.

Basis for the Induction

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

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:
 * $\displaystyle \left({x_1 x_2 \cdots x_m}\right)^{1/m} \le \frac 1 m \left({x_1 + x_2 + \cdots + x_m}\right)$

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

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


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

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

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