$(a+1)(b+1)(c+1)\leq4$ for triangle sides $a,b,c$ with $ab+bc+ac=1$ 
Given that $a,b,c$ are the lengths of the three sides of a triangle, and $ab+bc+ac=1$, the question is to prove $$(a+1)(b+1)(c+1)\leq4\,.$$

Any idea or hint would be appreciated.
This is Problem 6 of Round 1 of the BMO (British Mathematical Olympiad) 2010/2011, as can be seen here.
Remark. This question has been self-answered. Nevertheless, any new approach is always welcome!

We need to prove $$abc+a+b+c\leq2$$ or
$$(abc+(a+b+c)(ab+ac+bc))^2\leq4(ab+ac+bc)^3$$ or
$$\prod_{cyc}(a(b+c-a)+bc)\geq0$$ and we are done!
We can get a last factoring by the following way.
For $ab+ac+bc=a^2$ we obtain:
$$(abc+(a+b+c)(ab+ac+bc))^2=(abc+(a+b+c)a^2)^2=a^2(a^2+ab+ac+bc)^2=$$
$$=(ab+ac+bc)(2(ab+ac+bc))^2=4(ab+ac+bc)^3$$ and since we work with symmetric polynomials, we got the needed factorization.
