Exercise 489Algebra › InequalitiesHigh schoolAdded 3 Nov 2026

The AM-GM inequality

#AM-GM#inequality#arithmetic and geometric means

Statement

Let $a,b\ge0$ . Prove that
$\sqrt{ab}\le\frac{a+b}{2},$
with equality if and only if $a=b$ .

Hint

Consider $(\sqrt{a}-\sqrt{b})^2\ge0$ .

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Direct proof. For $a,b\ge0$ ,
$(\sqrt{a}-\sqrt{b})^2=a-2\sqrt{ab}+b\ge0,$
so $2\sqrt{ab}\le a+b$ , i.e. $\sqrt{ab}\le\frac{a+b}{2}$ .

Equality case. $(\sqrt{a}-\sqrt{b})^2=0$ iff $\sqrt{a}=\sqrt{b}$ , i.e. $a=b$ .

Conclusion. The geometric mean is at most the arithmetic mean, with equality iff the numbers are equal.