TheoremProved

AM–GM inequality

Statement

For non-negative real numbers $a_1,\dots,a_n$ , $\dfrac{a_1+a_2+\cdots+a_n}{n}\ge \sqrt[n]{a_1a_2\cdots a_n}$ , with equality iff $a_1=a_2=\cdots=a_n$ .

Why is it true?

Among all rectangles with a fixed perimeter, the square encloses the most area — spreading the side lengths apart to make them unequal only shrinks the product while the sum (perimeter) stays fixed. The AM–GM inequality is the same phenomenon in $n$ dimensions: for a fixed sum, the product of the numbers is largest exactly when they are all equal, and any imbalance among them can only make the product smaller.

Proof sketch

Cauchy's forward–backward induction proves it first for $n=2^k$ by repeatedly applying the two-variable case $\tfrac{a+b}{2}\ge\sqrt{ab}$ (itself equivalent to $(\sqrt a-\sqrt b)^2\ge 0$ ), then descends from $n=2^k$ to every smaller $n$ by applying the $2^k$ -variable inequality to the $n$ numbers padded with copies of their own average.

Proved by

proved 1821Augustin-Louis Cauchy

Topics that use this theorem

Inequalities

Related theorems

Cauchy–Schwarz inequality

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

G. H. Hardy, J. E. Littlewood, G. Pólya (1934). Inequalities