<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] Solve for $c$:
$$c=\frac{-b\pm\sqrt{12-3b^2}}{2}$$
To prove the inequality, we need to prove it for the larger of the two values of $c$ (the one with the sign "plus"):
$$b+c=\frac{b+\sqrt{12-3b^2}}{2}\le 2$$
i.e.
$$\sqrt{12-3b^2}\le 4-b$$
which is defined for $|b|\le 2$ - and in that interval the RHS is positive so we can square both sides:
$$12-3b^2\le 16-8b+b^2$$
or: $4b^2-8b+4=4(b-1)^2\ge 0$, which is obviously true. The equality is satisfied for $b=1$, which gives $c=1$.
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