Show if $a^2+b^2 \le 2$ then $a+b \le 2$ If $a^2+b^2 \le 2$ then show that $a+b \le2$
I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result

Yet another method, inspired by looking at the problem geometrically (try drawing the region $a^2+b^2\leq2$ and the line $a+b=2$): let $s=a+b$, $t=a-b$.  Then $a=\frac12(s+t)$ and $b=\frac12(s-t)$, so $a^2+b^2=\frac14\bigl((s^2+2st+t^2)+(s^2-2st+t^2)\bigr) = \frac12(s^2+t^2)$ and $a+b=s$, so the problem becomes: 

if $s^2+t^2\leq 4$ then show that $s\leq 2$.

But this is manifestly true; $t^2\geq 0$, so if $s^2+t^2\leq 4$ then $s^2\leq 4$ and so $s\leq 2$.
