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

Hint Use Cauchy-Schwarz.
Second solution
$$(a+b)^2=a^2+b^2+2ab \leq 2+2ab$$
You got that far, you are almost there:
By AM-GM
$$\sqrt{a^2b^2} \leq \frac{a^2+b^2}{2}$$
which implies $$2ab \leq a^2+b^2 \leq 2$$
