Dimostrare $x^2+y^2+z^2\ge xy+yz+zx$

Text

Prove that, for every triple of real numbers $x,y,z$, $x^2+y^2+z^2\ge xy+yz+zx.$

Solution

Consider twice the difference between the two sides and complete the squares: $2\big(x^2+y^2+z^2\big)-2\big(xy+yz+zx\big)=(x-y)^2+(y-z)^2+(z-x)^2.$ The sum of three squares is $\ge 0$, hence $x^2+y^2+z^2-(xy+yz+zx)\ge 0$. Equality holds if and only if $x-y=y-z=z-x=0$, i.e. $x=y=z$. $\boxed{x^2+y^2+z^2\ge xy+yz+zx,\qquad =\ \text{iff } x=y=z.}$