Cauchy's Mean Theorem/Proof of Equality Condition

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:
 * $A_n = G_n$


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

That is, all terms are equal.

Then:

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

Then:

So:
 * $A_n = G_n = n$

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:

which is trivially true.

Basis for the Induction
$\map P 2$ is the case:

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:

Then:

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:

But:
 * $\paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1 / k}$ is the geometric mean of $\set {x_1, x_2, \ldots, x_{k - 1}, G_{k - 1} }$
 * $\paren {\dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k$ is the geometric 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 $n$.
 * $\forall n \in \N: A_n = G_n \implies \forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$