<s>[INST] If $b^2+c^2+bc=3$ then $b+c\leq 2$ Suppose that $b,c\geq 0$ such that $b^2+c^2+bc=3$. Prove that $b+c\leq 2$.
I tried to do that by contradiction but I failed.
Indeed, if $b+c>2$ then $b^2+2bc+c^2>4$ then $(b^2+bc+c^2)+bc>4$. Hence $3+bc>4$ or equivalently $bc>1$.
 [/INST] By AM-GM $$b+c=\sqrt{(b+c)^2}=\sqrt{b^2+2bc+c^2}=\sqrt{b^2+\frac{4}{3}bc+c^2+\frac{2}{3}bc}\leq$$
$$\leq\sqrt{b^2+\frac{4}{3}bc+c^2+\frac{b^2+c^2}{3}}=\sqrt{\frac{4}{3}(b^2+bc+c^2)}=2.$$
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