## The AM-GM Inequality

### For positive numbers, the arithmetic mean is greater than or equal to the
geometric mean

**Proof using Jensen's Inequality:**

Jensen's Inequality is

∑(f(x_{i}))/n ≤ f(∑(x_{i})/n)
if f is concave (i.e. f''(x)<0, which is sometimes called "concave down"),

∑(f(x_{i}))/n ≥ f(∑(x_{i})/n)
if f is convex (i.e. f''(x)>0, which is sometimes called "concave up"),

with equality iff x_{1} = x_{2} = ... = x_{n}.

ln(x) is concave, so by Jensen's Inequality,

(ln(a_{1}) + ln(a_{2}) + ... + ln(a_{n}))/n ≤ ln((a_{1}+a_{2}+...+a_{n})/n),

Taking the exponential function of both sides,

geometric mean(a_{1},a_{2},...,a_{n}) ≤ arithmetic
mean(a_{1},a_{2},...,a_{n})

### Weighted AM-GM Inequality

Given n positive reals a_{1},a_{2},...,a_{n}, and n
weights w_{1},w_{2},...,w_{n} such that 0≤w_{i}≤1
and ∑w_{i}=1,

w_{1}a_{1}+w_{2}a_{2}+...+w_{n}a_{n}
≥ a_{1}^{w}1 a_{2}^{w}2
... a_{n}^{w}n

**Proof Outline:** same as above, using the weighted version of Jensen's
Inequality; or use the ordinary AM-GM inequality for rational weights (put all
the weights over a common denominator, d, and then show that the AM of the d
numbers (many identical) is never less than the GM of those same numbers); and
then extend to irrational weights by continuity.

### Internet references

Mathworld
-- Jensen's Inequality

### Related pages in this website

The Cauchy-Schwarz inequality

The Triangle Inequality

The Chebyshev Sum Inequality

Puzzles

The Quadratic Formula

The webmaster and author of this Math Help site is
Graeme McRae.