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:
- $A_n = G_n$
- $\forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$
That is, if and only if all terms are equal.
Then:
Proof
Necessary Condition
Let:
- $\forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j = x$
Then:
\(\ds A_n\) | \(=\) | \(\ds \dfrac 1 n \sum_{j \mathop = 1}^n x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 n n x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x\) |
\(\ds G_n\) | \(=\) | \(\ds \paren {\prod_{j \mathop = 1}^n x}^{\frac 1 n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x^n}^{\frac 1 n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x\) |
So:
- $A_n = G_n = x$
$\Box$
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:
\(\ds A_1\) | \(=\) | \(\ds G_1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 1 \sum_{j \mathop = 1}^1 x_j\) | \(=\) | \(\ds \paren {\prod_{j \mathop = 1}^1 x_j}^{1/1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_1\) | \(=\) | \(\ds x_1\) |
which is trivially true.
Basis for the Induction
$\map P 2$ is the case:
\(\ds A_2\) | \(=\) | \(\ds G_2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {x_1 + x_2} 2\) | \(=\) | \(\ds \sqrt {x_1 x_2}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\dfrac {x_1 + x_2} 2}^2\) | \(=\) | \(\ds x_1 x_2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_1^2 + 2 x_1 x_2 + x_2^2\) | \(=\) | \(\ds 4 x_1 x_2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_1^2 - 2 x_1 x_2 + x_2^2\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x_1 - x_2}^2\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_1\) | \(=\) | \(\ds x_2\) |
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 Forward-Backward 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:
- $A_k' := \ds \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j = y$
and:
- $G_k' := \ds \paren {\prod_{j \mathop = k + 1}^{2 k} }^{1 / k} x_j = y$
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:
- $A_k' := \ds \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j = y$
and:
- $G_k' := \ds \paren {\prod_{j \mathop = k + 1}^{2 k} }^{1 / k} x_j = y$
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:
\(\ds A_{2 k}\) | \(=\) | \(\ds \dfrac 1 {2 k} \sum_{j \mathop = 1}^{2 k} x_j\) | Definition of Arithmetic Mean | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {2 k} \paren {\sum_{j \mathop = 1}^k x_j + \sum_{j \mathop = k + 1}^{2 k} x_j}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {\dfrac 1 k \sum_{j \mathop = 1}^k x_j + \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {A_k + A_k'}\) | Definition of Arithmetic Mean | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {x + y}\) | Definition of $A_k$ and $A_k'$ |
\(\ds G_{2 k}\) | \(=\) | \(\ds \paren {\prod_{j \mathop = 1}^{2 k} x_j}^{1 / 2 k}\) | Definition of Geometric Mean | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {\prod_{j \mathop = 1}^{2 k} x_j}^{1 / k} }^{1 / 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {\prod_{j \mathop = 1}^k x_j}^{1 / k} \times \paren {\prod_{j \mathop = k + 1}^{2 k} x_j}^{1 / k} }^{1 / 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {G_k G_k'}^{1 / 2}\) | Definition of Geometric Mean | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x y}^{1 / 2}\) | Definition of $G_k$ and $G_k'$ |
Then:
\(\ds A_{2 k}\) | \(=\) | \(\ds G_{2 k}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 2 \paren {x + y}\) | \(=\) | \(\ds \paren {x y}^{1 / 2}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y\) | 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:
\(\ds G_{k - 1}\) | \(=\) | \(\ds A_{k - 1}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {k - 1} G_{k - 1}\) | \(=\) | \(\ds \paren {k - 1} A_{k - 1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds k G_{k - 1}\) | \(=\) | \(\ds \paren {k - 1} A_{k - 1} + G_{k - 1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {G_{k - 1}^{k - 1} G_{k - 1} }^{1/k}\) | \(=\) | \(\ds \dfrac {\paren {k - 1} A_{k - 1} + G_{k - 1} } k\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1 / k}\) | \(=\) | \(\ds \dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1 / k}\) | \(=\) | \(\ds \dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k\) |
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 Forward-Backward Induction.
Therefore $A_n \ge G_n$ for all $n$.
- $\forall n \in \N: A_n = G_n \implies \forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$
$\blacksquare$