Cauchy's Mean Theorem/Proof 1

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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 $A_n \ge G_n$.


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 $n$.

$\blacksquare$


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