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 $bc>1$.

We have
$$3 = b^2+bc+c^2 = \frac{3(b+c)^2+(b-c)^2}{4} \geqslant \frac{3}{4}(b+c)^2.$$
Therefore $(b+c)^2 \leqslant 4,$ so $b+c \leqslant 2.$
