It`s just Schur inequality .

schur gives: $ab+ac+bc \le \frac{1}{4}+\frac{9}{4}abc$ and by AM-GM: $ab+ac+bc \ge 3(abc)^{2/3} \ge 9abc$

it is very easy and famous inqeuality!) how they give you such a problms in national olympiad!)

The problem is a direct consequence of Schur's Inequality, and its purpose is just to get familiar with this useful inequality. So, don't say something like this if you are not well-informed about Moroccan's Contests mister arshakus If i were in your shoes, i would take a look at this : http://www.artofproblemsolving.com/Forum/resources.php?c=205&cid=199&year=2011

Let $a \ge b \ge c$. From MA - MG, $a(b+c) \le(\frac{a + (b+c)}{2})^2 = \frac{1}{4}$. Also from $a+b+c = 1$ we have that $3a \ge 1 => 3abc \ge bc$. Hence $a(b+c) + bc = ab+bc+ac \le \frac{1}{4} + 3abc$. I did the other part with just AM- GM. What version of schur are you using?

I'm talking about this version : $\sum_{cyc} a(a-b)(a-c) \ge 0$ $\Leftrightarrow (a+b+c)^3-4(a+b+c)(ab+bc+ca)+9abc \ge 0$