<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] Let $a=\sqrt{2}\cos\theta$, $y=\sqrt{2}\sin\theta$. Then $a^2+b^2=2$, and $a+b=\sqrt{2}(\cos\theta+\sin\theta)$, which is a maximum at $\theta=\frac{\pi}{4}$, at which case $a+b=2$. So $a+b\le 2$.
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