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 $bc>1$.
 A: $ 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.
A: By AM-GM $$b+c=\sqrt{(b+c)^2}=\sqrt{b^2+2bc+c^2}=\sqrt{b^2+\frac{4}{3}bc+c^2+\frac{2}{3}bc}\leq$$
$$\leq\sqrt{b^2+\frac{4}{3}bc+c^2+\frac{b^2+c^2}{3}}=\sqrt{\frac{4}{3}(b^2+bc+c^2)}=2.$$
A: 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 $b+c \leqslant 2.$
A: 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 $c=1$.
A: 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!
A: 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., $$(-k)^2\ge4(k^2-3)\iff k^2\le4\iff-2\le k\le2$$
