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.

$$ \begin{align}&  a^2+b^2+c^2+ab+ac+bc\le1 \\ 
& (a+b+c)^2 -(ab+bc+ac) \le 1 \\ 
& (a+b+ c) \le \sqrt{1+(ab+bc+ac)} \quad \quad \color{red}{\text{ (1.)}} \end{align}$$
Also note that :
$$\frac {(ab+bc+ac)}{3} \ge \sqrt[3]{a^2b^2c^2}$$
Assuming you are correct :
$${(ab+bc+ac)} \ge 3 \cdot\frac16 \implies (ab+bc+ac) \ge \frac12$$
Hence from $\color{red}{\text{ (1.)}}$:
$$(a+b+c) \le \sqrt{1+\frac12} \implies \boxed {\color {blue}{(a+b+c) \le \sqrt {\frac32}}}$$
