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.