Cauchy's Mean Theorem

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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:

$\displaystyle A_n = \frac 1 n \left({\sum_{k \mathop = 1}^n x_k}\right)$


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

$\displaystyle G_n = \left({\prod_{k \mathop = 1}^n x_k}\right)^{1/n}$


We prove the result by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:

For all positive real numbers $x_1, x_2, \ldots, x_n: A_n \ge G_n$.


$P(1)$ is true, as this just says:

$\dfrac {x_1} 1 \ge x_1^{1/1}$

which is trivially true.


Basis for the Induction

$P(2)$ 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:

\(\displaystyle 0\) \(\le\) \(\displaystyle \left({\sqrt{x_1} - \sqrt{x_2} }\right)^2\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x_1 - 2\sqrt{x_1 x_2} + x_2\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \sqrt{x_1 x_2}\) \(\le\) \(\displaystyle \frac {x_1 + x_2} 2\) $\quad$ $\quad$

This is our basis for the induction.


Induction Hypothesis

Now we show that:

$(1): \quad$ If $P \left({2^k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({2^{k+1}}\right)$ is true
$(2): \quad$ If $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k-1}\right)$ 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 $2^{k+1} = 2m$.

Since $P \left({m}\right)$ is true:

$\displaystyle \left({x_1 x_2 \cdots x_m}\right)^{1/m} \le \frac 1 m \left({x_1 + x_2 + \cdots + x_m}\right)$

Also:

$\displaystyle \left({x_{m+1} x_{m+2} \cdots x_{2m}}\right)^{1/m} \le \frac 1 m \left({x_{m+1} + x_{m+2} + \cdots + x_{2m}}\right)$

But we have $P(2)$, so:

$\displaystyle \left({\left({x_1 x_2 \cdots x_m}\right)^{1/m} \left({x_{m+1} x_{m+2} \cdots x_{2m}}\right)^{1/m}}\right)^{1/2} \le \frac 1 2 \left({\frac {x_1 + x_2 + \cdots + x_m} m + \frac {x_{m+1} + x_{m+2} + \cdots + x_{2m}} m}\right)$

So:

$\displaystyle \left({x_1 x_2 \cdots x_{2m}}\right)^{1/2m} \le \frac {x_1 + x_2 + \cdots + x_{2m}} {2m}$

So $P \left({2m}\right) = P \left({2^{k+1}}\right)$ holds.

So $P \left({2^n}\right)$ holds for all $n$ by induction.


Now suppose $P \left({k}\right)$ holds. Then:

\(\displaystyle \left({x_1 x_2 \cdots x_{k-1} G_{k-1} }\right)^{1/k}\) \(\le\) \(\displaystyle \frac {x_1 + x_2 + \cdots + x_{k-1} + G_{k-1} } k\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({G_{k-1}^{k-1} G_{k-1} }\right)^{1/k}\) \(\le\) \(\displaystyle \frac {\left({k-1}\right) A_{k-1} + G_{k-1} } k\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle k G_{k-1}\) \(\le\) \(\displaystyle \left({k-1}\right) A_{k-1} + G_{k-1}\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({k-1}\right) G_{k-1}\) \(\le\) \(\displaystyle \left({k-1}\right) A_{k-1}\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle G_{k-1}\) \(\le\) \(\displaystyle A_{k-1}\) $\quad$ $\quad$

So $P \left({k}\right) \implies P \left({k-1}\right)$ and the result follows by Backwards Induction.


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

$\blacksquare$


Also known as

It is widely known as the Arithmetic Mean-Geometric Mean Inequality or AM-GM Inequality.

Some sources give this as Cauchy's formula.


Also see


Source of Name

This entry was named for Augustin Louis Cauchy.


Sources