Com10atorics wrote: If $a>1,b>1$ are the legths of the catheti of an right triangle and $c$ the length of its hypotenuse, prove that $a+b\leq c\sqrt 2$ \[ a+b=\sqrt{(a+b)^2}\leq\sqrt{(a+b)^2+(a-b)^2}=\sqrt{2(a^2+b^2)}=c\sqrt{2} \]

The condition that $a, b> 1$ is unnecessary. Since $c=\sqrt{a^2+b^2}$ then $$a+b\leqslant c\sqrt{2} \Leftrightarrow (a-b)^2 \geqslant 0$$Equality occur if and only if $a=b$, i.e. triangle is isosceles.

$a+b= \sqrt{(a+b)^2} \leq \sqrt{(a+b)^2+(a-b)^2} = \sqrt{a^2+2ab+b^2+a^2-2ab+b^2} = \sqrt{2(a^2+b^2)} =c \sqrt{2}$ $\blacksquare$