<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] Suffices to show if $a^2+b^2 = 2$ then $a+b \leq 2$. From the constraint consider
$$
f(a) = a + \sqrt{2-a^2}
$$
and we need to prove $f(a) \leq 2$ over $[0,\sqrt{2}]$.
$$
f'(a) = 1 - \frac{a}{\sqrt{2-a^2}} \Leftrightarrow a = 1
$$
which is a maximum by 1st derivative test.
Since $f(0), f(1), f(\sqrt{2})$ are $\sqrt{2}, 2, \sqrt{2}$ we have $f(a) \leq 2$ as desired.
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