Cauchy's Mean Theorem/Proof 2

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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

Let:

$\map f x = \ln x$

for $x > 0$.

With a view to apply Jensen's Inequality: Real Analysis: Corollary, we can show that $f$ is concave on $\openint 0 \infty$.

By Second Derivative of Concave Real Function is Non-Positive, it is sufficient to show that $\map {f''} x \le 0$ for all $x > 0$.

We have, by Derivative of Natural Logarithm:

$\map {f'} x = \dfrac 1 x$

We then have, by Derivative of Power:

$\map {f''} x = -\dfrac 1 {x^2}$

As $x^2 > 0$ for all $x > 0$, we have:

$\dfrac 1 {x^2} > 0$

Therefore:

$-\dfrac 1 {x^2} = \map {f''} x < 0$

so $f$ is indeed concave on $\openint 0 \infty$.

As $x_1, x_2, \ldots, x_n$ are all positive, they all lie in the interval $\openint 0 \infty$.

We therefore have, by Jensen's Inequality: Real Analysis: Corollary:

$\ds \map \ln {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \ge \frac {\sum_{k \mathop = 1}^n \lambda_k \map \ln {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$

for real $\lambda_1, \lambda_2, \ldots, \lambda_n \ge 0$, with at least one of which being non-zero.

As $n$ is a positive integer, we have:

$\dfrac 1 n > 0$

We can therefore set:

$\lambda_i = \dfrac 1 n$

for $1 \le i \le n$.

This gives:

\(\ds \map \ln {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} }\) \(=\) \(\ds \map \ln {\frac {\sum_{k \mathop = 1}^n x_k} {n \sum_{k \mathop = 1}^n \frac 1 n} }\)
\(\ds \) \(=\) \(\ds \map \ln {\frac {\sum_{k \mathop = 1}^n x_k} n}\)
\(\ds \) \(=\) \(\ds \map \ln {A_n}\) Definition of Arithmetic Mean

and:

\(\ds \frac {\sum_{k \mathop = 1}^n \lambda_k \map \ln {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}\) \(=\) \(\ds \frac {\sum_{k \mathop = 1}^n \map \ln {x_k} } {n \sum_{k \mathop = 1}^n \frac 1 n}\)
\(\ds \) \(=\) \(\ds \frac {\sum_{k \mathop = 1}^n \map \ln {x_k} } n\)
\(\ds \) \(=\) \(\ds \frac 1 n \map \ln {\prod_{k \mathop = 1}^n x_k}\) Sum of Logarithms
\(\ds \) \(=\) \(\ds \map \ln {\paren {\prod_{k \mathop = 1}^n x_k}^{1/n} }\) Logarithm of Power
\(\ds \) \(=\) \(\ds \map \ln {G_n}\) Definition of Geometric Mean

We therefore have:

$\map \ln {A_n} \ge \map \ln {G_n}$

Note that for $x > 0$:

$\dfrac 1 x = \map {f'} x > 0$

Therefore, by Derivative of Monotone Function, $f$ is increasing on $\openint 0 \infty$.

We therefore have:

$A_n \ge G_n$

$\blacksquare$


Proof of Equality Condition

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 $n$.

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

$\blacksquare$