AM-HM Inequality

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Theorem

Let $x_1, x_2, \ldots, x_n \in \R_{> 0}$ be strictly positive real numbers.

Let $A_n $ be the arithmetic mean of $x_1, x_2, \ldots, x_n$.

Let $H_n$ be the harmonic mean of $x_1, x_2, \ldots, x_n$.

Then $A_n \ge H_n$.

Proof $1$

$A_n$ is defined as:

$\ds A_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n x_k}$

$H_n$ is defined as:

$\ds \frac 1 {H_n} = \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {x_k} }$

We have that:

$\forall k \in \closedint 1 n: x_k > 0$

From Positive Real has Real Square Root, we can express each $x_k$ as a square:

$\forall k \in \closedint 1 n: x_k = y_k^2$

without affecting the result.

Thus we have:

$\ds A_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n y_k^2}$

$\ds \frac 1 {H_n} = \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {y_k^2} }$

Multiplying $A_n$ by $\dfrac 1 {H_n}$:

$\ds \frac {A_n} {H_n}$

$=$

$\ds \frac 1 n \paren {\sum_{k \mathop = 1}^n y_k^2} \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {y_k^2} }$

$\ds $

$\ge$

$\ds \frac 1 {n^2} \paren {\sum_{k \mathop = 1}^n \frac {y_k} {y_k} }^2$

Cauchy's Inequality

$\ds $

$=$

$\ds \frac 1 {n^2} \paren {\sum_{k \mathop = 1}^n 1}^2$

$\ds $

$=$

$\ds \frac {n^2} {n^2} = 1$

So:

$\dfrac {A_n} {H_n} \ge 1$

and so from Real Number Axiom $\R \text O2$: Usual Ordering is Compatible with Multiplication:

$A_n \ge H_n$

$\blacksquare$

Proof $2$

By Hölder Mean for Exponent 1 is Arithmetic Mean:

$A_n = \dfrac {x_1 + x_2 + \cdots + x_n} n = \map {M_1} {x_1, x_2, \ldots, x_n}$

By Hölder Mean for Exponent -1 is Harmonic Mean:

$H_n = \dfrac 1 {\dfrac 1 n \paren {\dfrac 1 {x_1} + \dfrac 1 {x_2} + \cdots + \dfrac 1 {x_n} } } = \map {M_{-1} } {x_1, x_2, \ldots, x_n}$

The result follows from Inequality of Hölder Means.

$\blacksquare$

Also see

Sources

1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.12 \ (5)$
2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): mean
2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): mean