<s>[INST] 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
 [/INST] First note that if $a=0$ or $b=0$, then the question is easy. So assume that both are non-zero. Consider the value $ab$. We can show that we must have $|ab|\le1$. 
Assume by contrary that we have:  $|ab| > 1$. This means $$\frac{1}{|b|}<|a|$$
Expanding: $$0\le(b^2-1)^2$$
we get:$$2\le \frac{1}{b^2}+b^2,$$
Thus, using the original ineqaulity we see:
$$
2\le \frac{1}{b^2}+b^2\lt a^2+b^2\le2
$$
which is a contradiction. Hence,$|ab|\le1$, and we can continue your original inequality, by writing:
$$
a+b\le \sqrt{2+2ab}\le \sqrt{2+2}=2
$$
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