Solution:
Let $a$ and $b$ be the legs of a right triangle and $c$ its hypotenuse. We know that
$$\begin{aligned} (a-b)^2 & \geq 0 \\ a^2 + b^2 & \geq 2ab \end{aligned}$$
By the Pythagorean Theorem, we have
$$\begin{aligned} (a+b)^2 & = a^2 + 2ab + b^2 \\ & \leq 2(a^2 + b^2) \\ & = 2c^2 \end{aligned}$$
Therefore,
$$a + b \leq \sqrt{2} c$$