Maximum value of $a+b+c$ in an inequality Given that $a$, $b$ and $c$ are real positive numbers, find the maximum possible value of $a+b+c$, if
$$a^2+b^2+c^2+ab+ac+bc\le1.$$
From the AM-GM theorem, I have
$$a^2+b^2+c^2+ab+ac+bc\geq 6\sqrt[6]{a^4b^4c^4} = 6\sqrt[3]{a^2b^2c^2} \\
6\sqrt[3]{a^2b^2c^2} \le1 \\
a^2b^2c^2 \le \frac{1}{216} \\
abc \le \frac{\sqrt{6}}{36}$$
However, I don't know where to go from here.

$$4(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2=6(a^2+b^2+c^2+ab+bc+ca)$$
and so $$4(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2\le 6$$
Each of $(a-b)^2,(b-c)^2,(c-a)^2$ is non-negative and so 
$$4(a+b+c)^2\le 6.$$
Therefore $a+b+c\le \sqrt \frac{3}{2}$.
