$\begin{array}{l}\sqrt{a^2 + b^2} \geq \dfrac{a+b}{\sqrt2}\\ \Leftrightarrow \sqrt{2(a^2 + b^2)} \geq a + b\\ \Leftrightarrow 2(a^2 + b^2) \geq (a + b)^2\\ \Leftrightarrow 2a^2 + 2b^2 \geq a^2 + 2ab + b^2\\ \Leftrightarrow a^2 - 2ab + b^2 \geq 0\\ \Leftrightarrow (a-b)^2 \geq 0\quad \text{(luôn đúng)}\\ Vậy\,\,\sqrt{a^2 + b^2} \geq \dfrac{a+b}{\sqrt2}\,\,\forall a,b \geq 0\end{array}$