Dimostrare $\sqrt{\dfrac{a^2+b^2}{2}}\ge\dfrac{a+b}{2}$

Text

Prove that the root-mean-square of two real numbers is never less than their arithmetic mean: for every $a,b$ with $a+b\ge 0$, $\sqrt{\frac{a^2+b^2}{2}}\ge\frac{a+b}{2}.$

Solution

Both sides are non-negative (the left one by definition of the root, the right one because $a+b\ge 0$), so we may compare the squares: $\frac{a^2+b^2}{2}-\left(\frac{a+b}{2}\right)^2=\frac{2a^2+2b^2-(a^2+2ab+b^2)}{4}=\frac{(a-b)^2}{4}\ge 0.$ Hence the square of the left side is $\ge$ the square of the right side, and by monotonicity of the root the claim follows. Equality iff $a=b$. $\boxed{\sqrt{\frac{a^2+b^2}{2}}\ge\frac{a+b}{2},\qquad =\ \text{iff } a=b.}$