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 $bc>1$.

Let $b+c=k\iff b=k-c$
$$3=b^2+bc+c^2=(k-c)^2+(k-c)c+c^2$$
$$\iff c^2-kc+k^2-3=0$$ which is a quadratic equation in $c$
As $c$ is real, the discriminant must be $\ge0$
i.e., $$(-k)^2\ge4(k^2-3)\iff k^2\le4\iff-2\le k\le2$$
