<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] $ 0 \leq (b-c)^{2} =b^{2}+c^{2}-2bc=3-3bc$ so $bc \leq 1$. Now $(b+c)^{2}=b^{2}+c^{2}+2bc=3+bc \leq  4 $ Now take square root.
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