AM–GM inequality

In mathematics, the inequality of arithmetic and geometric means, or more briefly the AM–GM inequality, states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list; and further, that the two means are equal if and only if every number in the list is the same (in which case they are both that number).

The simplest non-trivial case is for two non-negative numbers $x$ and $y$ , that is,

{\frac {x+y}{2}}\geq {\sqrt {xy}}

with equality if and only if $x = y$ . This follows from the fact that the square of a real number is always non-negative (greater than or equal to zero) and from the identity $(a \pm b) 2 = a 2 \pm 2 ab + b 2$ :

{\begin{aligned}0&\leq (x-y)^{2}\\&=x^{2}-2xy+y^{2}\\&=x^{2}+2xy+y^{2}-4xy\\&=(x+y)^{2}-4xy.\end{aligned}}

Hence $(x + y) 2 \geq 4 xy$ , with equality when $(x - y) 2 = 0$ , i.e. $x = y$ . The AM–GM inequality then follows from taking the positive square root of both sides and then dividing both sides by 2.

For a geometrical interpretation, consider a rectangle with sides of length $x$ and $y$ ; it has perimeter $2 x + 2 y$ and area $xy$ . Similarly, a square with all sides of length $\sqrt xy$ has the perimeter $4 \sqrt xy$ and the same area as the rectangle. The simplest non-trivial case of the AM–GM inequality implies for the perimeters that $2 x + 2 y \geq 4 \sqrt xy$ and that only the square has the smallest perimeter amongst all rectangles of equal area.

The simplest case is implicit in Euclid's Elements, Book V, Proposition 25.

Extensions of the AM–GM inequality treat weighted means and generalized means.