\(\sqrt{a^2+b^2}\ge\dfrac{a+b}{\sqrt{2}}\left(1\right)\)
\(\Leftrightarrow\sqrt{2}.\sqrt{a^2+b^2}\ge a+b\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow2a^2+2b^2>a^2+b^2+2ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) (2)
(2) đúng => (1) đúng
-----------------------GOOD LUCK----------------------