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

$A_n$ 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:

$A_n \ge H_n$

$\blacksquare$


Also see


Sources

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