# Cauchy's Mean Theorem/Proof 1

## 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:

$A_n \ge G_n$

with equality holding if and only if:

$\forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$

That is, if and only if all terms are equal.

## Proof

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

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

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

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

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:

 $\displaystyle 0$ $\le$ $\displaystyle \paren {\sqrt {x_1} - \sqrt {x_2} }^2$ $\displaystyle$ $=$ $\displaystyle x_1 - 2\sqrt {x_1 x_2} + x_2$ $\displaystyle \leadsto \ \$ $\displaystyle \sqrt {x_1 x_2}$ $\le$ $\displaystyle \frac {x_1 + x_2} 2$

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 Backwards 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:

 $\displaystyle \paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1/k}$ $\le$ $\displaystyle \dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {G_{k - 1}^{k - 1} G_{k - 1} }^{1/k}$ $\le$ $\displaystyle \dfrac {\paren {k - 1} A_{k - 1} + G_{k - 1} } k$ $\displaystyle \leadsto \ \$ $\displaystyle k G_{k - 1}$ $\le$ $\displaystyle \paren {k - 1} A_{k - 1} + G_{k - 1}$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {k - 1} G_{k - 1}$ $\le$ $\displaystyle \paren {k - 1} A_{k - 1}$ $\displaystyle \leadsto \ \$ $\displaystyle G_{k - 1}$ $\le$ $\displaystyle A_{k - 1}$

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

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

$\blacksquare$

## Proof of Equality Condition

### Necessary Condition

Let:

$\forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j = x$

Then:

 $\displaystyle A_n$ $=$ $\displaystyle \dfrac 1 n \sum_{j \mathop = 1}^n x$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 n n x$ $\displaystyle$ $=$ $\displaystyle x$

 $\displaystyle G_n$ $=$ $\displaystyle \paren {\prod_{j \mathop = 1}^n x}^{\frac 1 n}$ $\displaystyle$ $=$ $\displaystyle \paren {x^n}^{\frac 1 n}$ $\displaystyle$ $=$ $\displaystyle x$

So:

$A_n = G_n = n$

$\Box$

### Sufficient Condition

Let $A_n = G_n$.

We prove the result by induction:

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:

$A_n = G_n \implies \forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$

$\map P 1$ is true, as this just says:

 $\displaystyle A_1$ $=$ $\displaystyle G_1$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac 1 1 \sum_{j \mathop = 1}^1 x_j$ $=$ $\displaystyle \paren {\prod_{j \mathop = 1}^1 x_j}^{1/1}$ $\displaystyle \leadsto \ \$ $\displaystyle x_1$ $=$ $\displaystyle x_1$

which is trivially true.

#### Basis for the Induction

$\map P 2$ is the case:

 $\displaystyle A_2$ $=$ $\displaystyle G_2$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {x_1 + x_2} 2$ $=$ $\displaystyle \sqrt {x_1 x_2}$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {\dfrac {x_1 + x_2} 2}^2$ $=$ $\displaystyle x_1 x_2$ $\displaystyle \leadsto \ \$ $\displaystyle x_1^2 + 2 x_1 x_2 + x_2^2$ $=$ $\displaystyle 4 x_1 x_2$ $\displaystyle \leadsto \ \$ $\displaystyle x_1^2 - 2 x_1 x_2 + x_2^2$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {x_1 - x_2}^2$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle x_1$ $=$ $\displaystyle x_2$

This is our basis for the induction.

#### Induction Hypothesis

Now we show that:

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

This is our first induction hypothesis:

$A_k = G_k \implies \forall i, j \in \set {1, 2, \ldots, k}: x_i = x_j = x$

Also, let:

$A_k' := \displaystyle \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j = y$

and:

$G_k' := \displaystyle \paren {\prod_{j \mathop = k + 1}^{2 k} }^{1 / k} x_j = y$

By the induction hypothesis:

$A_k' = G_k' \implies \forall i, j \in \set {k + 1, k + 2, \ldots, 2 k}: x_i = x_j = y$

We need to show:

$A_{2 k} = G_{2 k} \implies \forall i, j \in \set {1, 2, \ldots, 2 k}: x_i = x_j = x$

#### Induction Step

This is our induction step:

Suppose:

$A_{2 k} = G_{2 k}$

Let:

$A_k' := \displaystyle \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j = y$

and:

$G_k' := \displaystyle \paren {\prod_{j \mathop = k + 1}^{2 k} }^{1 / k} x_j = y$

By the induction hypothesis:

$A_k' = G_k' \implies \forall i, j \in \set {k + 1, k + 2, \ldots, 2 k}: x_i = x_j = y$

Then:

 $\displaystyle A_{2 k}$ $=$ $\displaystyle \dfrac 1 {2 k} \sum_{j \mathop = 1}^{2 k} x_j$ Definition of Arithmetic Mean $\displaystyle$ $=$ $\displaystyle \dfrac 1 {2 k} \paren {\sum_{j \mathop = 1}^k x_j + \sum_{j \mathop = k + 1}^{2 k} x_j}$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 2 \paren {\dfrac 1 k \sum_{j \mathop = 1}^k x_j + \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j}$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 2 \paren {A_k + A_k'}$ Definition of Arithmetic Mean $\displaystyle$ $=$ $\displaystyle \dfrac 1 2 \paren {x + y}$ Definition of $A_k$ and $A_k'$

 $\displaystyle G_{2 k}$ $=$ $\displaystyle \paren {\prod_{j \mathop = 1}^{2 k} x_j}^{1 / 2 k}$ Definition of Geometric Mean $\displaystyle$ $=$ $\displaystyle \paren {\paren {\prod_{j \mathop = 1}^{2 k} x_j}^{1 / k} }^{1 / 2}$ $\displaystyle$ $=$ $\displaystyle \paren {\paren {\prod_{j \mathop = 1}^k x_j}^{1 / k} \times \paren {\prod_{j \mathop = k + 1}^{2 k} x_j}^{1 / k} }^{1 / 2}$ $\displaystyle$ $=$ $\displaystyle \paren {G_k G_k'}^{1 / 2}$ Definition of Geometric Mean $\displaystyle$ $=$ $\displaystyle \paren {x y}^{1 / 2}$ Definition of $G_k$ and $G_k'$

Then:

 $\displaystyle A_{2 k}$ $=$ $\displaystyle G_{2 k}$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac 1 2 \paren {x + y}$ $=$ $\displaystyle \paren {x y}^{1 / 2}$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle y$ Basis for the Induction

That is:

$\forall i, j \in \set {1, 2, \ldots, k}: x_i = x_j = x$
$\forall i, j \in \set {k + 1, k + 2, \ldots, 2 k}: x_i = x_j = x$

Hence:

$\forall i, j \in \set {1, 2, \ldots, 2 k}: x_i = x_j = x$

Now suppose $\map P k$ holds.

Then:

 $\displaystyle G_{k - 1}$ $=$ $\displaystyle A_{k - 1}$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {k - 1} G_{k - 1}$ $=$ $\displaystyle \paren {k - 1} A_{k - 1}$ $\displaystyle \leadsto \ \$ $\displaystyle k G_{k - 1}$ $=$ $\displaystyle \paren {k - 1} A_{k - 1} + G_{k - 1}$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {G_{k - 1}^{k - 1} G_{k - 1} }^{1/k}$ $=$ $\displaystyle \dfrac {\paren {k - 1} A_{k - 1} + G_{k - 1} } k$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1 / k}$ $=$ $\displaystyle \dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1 / k}$ $=$ $\displaystyle \dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k$

But:

$\paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1 / k}$ is the geometric mean of $\set {x_1, x_2, \ldots, x_{k - 1}, G_{k - 1} }$
$\dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k$ is the arithmetic mean of $\set {x_1, x_2, \ldots, x_{k - 1}, G_{k - 1} }$

We have that $\set {x_1, x_2, \ldots, x_{k - 1}, G_{k - 1} }$ has $k$ elements.

Hence by the induction hypothesis:

$\forall i, j \in \set {1, 2, \ldots, k - 1}: x_i = x_j = G_{k - 1}$

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

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

$\forall n \in \N: A_n = G_n \implies \forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$

$\blacksquare$