Dimostrare $\sqrt{uv}\le\dfrac{u+v}{2}$ per $u,v\ge 0$

Text

Prove the inequality between the geometric and arithmetic mean: for every pair of numbers $u,v\ge 0$, $\sqrt{uv}\le\frac{u+v}{2}.$

Solution

Since $u,v\ge 0$, we can write $u=(\sqrt{u})^2$ and $v=(\sqrt{v})^2$. Compute the difference: $\frac{u+v}{2}-\sqrt{uv}=\frac{u-2\sqrt{uv}+v}{2}=\frac{(\sqrt{u}-\sqrt{v})^2}{2}\ge 0.$ Hence $\dfrac{u+v}{2}\ge\sqrt{uv}$. Equality holds if and only if $\sqrt{u}=\sqrt{v}$, i.e. $u=v$. $\boxed{\sqrt{uv}\le\frac{u+v}{2},\qquad =\ \text{iff } u=v.}$