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.

Hint: the inequality is equivalent to 
$$
(a+b)^{2} + (b+c)^{2} + (c+a)^{2} \leq 2.
$$
Now put $x = b+c, y = c+a, z = a +b$ and then we need to find maximum of $a + b + c  = \frac{1}{2}(x+y+z)$ under the condition $x^{2} + y^{2} + z^{2} \leq 2$ and you may use Cauchy-Schwartz inequality. Note that you need to check $x + y >z, y +z>x, z+x>y$ for $(x, y, z)$ that gives maximum. 
