Ta có $(a-b)^2 \geq 0\Rightarrow a^2-2ab+b^2 \geq 0\Rightarrow a^2+b^2 \geq2ab\Rightarrow 2(a^2+b^2) \geq a^2+2ab+b^2 \Rightarrow 2(a^2+b^2) \geq(a+b)^2\Rightarrow \sqrt{2(a^2+b^2)} \geq a+b$mà $a^2+b^2=c^2\Rightarrow \sqrt{2c^2} \geq a+b\Rightarrow a+b\leq \sqrt{2}.c$ (đpcm)
Dấu bằng xảy ra khi $a=b= \frac{\sqrt{2}}{2}.c$