# Cauchy's Mean Theorem

## Contents

## 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 1

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 $n$.

$\blacksquare$

## Proof 2

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:

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

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:

\(\displaystyle \map \ln {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} }\) | \(=\) | \(\displaystyle \map \ln {\frac {\sum_{k \mathop = 1}^n x_k} {n \sum_{k \mathop = 1}^n \frac 1 n} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \ln {\frac {\sum_{k \mathop = 1}^n x_k} n}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \ln {A_n}\) | Definition of Arithmetic Mean |

and:

\(\displaystyle \frac {\sum_{k \mathop = 1}^n \lambda_k \map \ln {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}\) | \(=\) | \(\displaystyle \frac {\sum_{k \mathop = 1}^n \map \ln {x_k} } {n \sum_{k \mathop = 1}^n \frac 1 n}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {\sum_{k \mathop = 1}^n \map \ln {x_k} } n\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 n \map \ln {\prod_{k \mathop = 1}^n x_k}\) | Sum of Logarithms | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \ln {\paren {\prod_{k \mathop = 1}^n x_k}^{1/n} }\) | Logarithm of Power | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \ln {G_n}\) | 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:

- $A_n \ge G_n$

$\blacksquare$

## 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.