49th Mathematical Olympiad in Ukraine

Using for the first denominator such inequality $3a^2+b^2+2ac=2a^2+a^2+b^2+2ac\ge 2a^2+2ab+2ac=2a(a+b+c)$ and analogously for the rest two denominators. Then the left-hand side becomes in the form of: $$\begin{array}{rl}\frac{a}{3a^2+b^2+2ac}+\frac{b}{3b^2+c^2+2ab}+\frac{c}{3c^2+a^2+2bc}& \le \frac{3}{2a(a+b+c)}+\frac{3}{2b(a+b+c)}+\frac{3}{2c(a+b+c)}=\frac{3}{2(a+b+c)}\text{that completes the proof.}\end{array}$$