Cauchy's Mean Theorem/Proof 1

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


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

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


 * $\ds G_n = \paren {\prod_{k \mathop = 1}^n x_k}^{1/n}$

We prove the result by induction:

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
 * For all positive real numbers $x_1, x_2, \ldots, x_n: A_n \ge G_n$.

$\map P 1$ is true, as this just says:
 * $\dfrac {x_1} 1 \ge {x_1}^{1/1}$

which is trivially true.

Basis for the Induction
$\map P 2$ 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:


 * $(1): \quad$ If $\map P {2^k}$ is true, where $k \ge 1$, then it logically follows that $\map P {2^{k + 1} }$ is true
 * $(2): \quad$ If $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k - 1}$ is true.

The result will follow by Forward-Backward Induction.

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

Because $\map P m$ is true:
 * $\paren {x_1 x_2 \dotsm x_m}^{1/m} \le \dfrac 1 m \paren {x_1 + x_2 + \dotsb + x_m}$

Also:
 * $\paren {x_{m + 1} x_{m + 2} \dotsm x_{2 m} }^{1/m} \le \dfrac 1 m \paren {x_{m + 1} + x_{m + 2} + \dotsb + x_{2 m} }$

But we have $\map P 2$, so:


 * $\paren {\paren {x_1 x_2 \dotsm x_m}^{1/m} \paren {x_{m + 1} x_{m + 2} \dotsm x_{2 m} }^{1/m} }^{1/2} \le \dfrac 1 2 \paren {\dfrac {x_1 + x_2 + \cdots + x_m} m + \dfrac {x_{m + 1} + x_{m + 2} + \dotsb + x_{2 m} } m}$

So:
 * $\paren {x_1 x_2 \dotsm x_{2 m} }^{1/2m} \le \dfrac {x_1 + x_2 + \dotsb + x_{2 m} } {2 m}$

So $\map P {2 m} = \map P {2^{k + 1} }$ holds.

So $\map P {2^n}$ holds for all $n$ by induction.

Now suppose $\map P k$ holds.

Then:

So $\map P k \implies \map P {k - 1}$ and the result follows by Forward-Backward Induction.

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