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 $\color{red}{\text{ (1.)}}$:
$$(a+b+c) \le \sqrt{1+\frac12} \implies \boxed {\color {blue}{(a+b+c) \le \sqrt {\frac32}}}$$
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.
Notice that $$ab+bc+ca\leq\frac{(a+b+c)^2}{3}.$$
Hence \begin{align} a^2+b^2+c^2+ab+bc+ca&=(a+b+c)^2-(ab+bc+ca)\\&\geq (a+b+c)^2-\frac{(a+b+c)^2}{3}\\&=\frac{2}{3}(a+b+c)^2. \end{align}
That is $$\frac{2}{3}(a+b+c)^2\leq 1,$$ which we get $$a+b+c\leq\sqrt{\frac{3}{2}}.$$
$$a+b+c=\sqrt{\sum_{cyc}(a^2+2ab)}\leq\sqrt{\sum_{cyc}\left(a^2+\frac{1}{2}a^2+\frac{3}{2}ab\right)}\leq\sqrt{\frac{3}{2}}.$$ The equality occurs for $a=b=c$ and $\sum\limits_{cyc}(a^2+ab)=1,$ which says that we got a maximal value.
$$4(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2=6(a^2+b^2+c^2+ab+bc+ca)$$ 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 $a+b+c\le \sqrt \frac{3}{2}$.