\(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\Rightarrow\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Rightarrow2a^2+2b^2-a^2-2ab-b^2\ge0\Rightarrow a^2-2ab+b^2\ge0\)
\(\Rightarrow\left(a-b\right)^2\ge0\) (luôn đúng) \(\Rightarrow\) đpcm