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 $bc>1$.

The Contradiction method also helps here.
Indeed, let $b+c>2,$ $b=kx$, $c=ky$, where $k>0$ and $x+y=2$.
Thus, $$k(x+y)>2,$$ which gives $k>1$ and
$$3=b^2+bc+c^2=k^2(x^2+xy+y^2)>x^2+xy+y^2,$$ which is a contradiction because we'll prove now that $$x^2+xy+y^2\geq3.$$
Indeed, we need to prove that:
$$x^2+xy+y^2\geq3\left(\frac{x+y}{2}\right)^2$$ or
$$(x-y)^2\geq0,$$ which is obvious.
Id est, the assumption that $b+c>2$ was wrong and we are done!
