Croatian Mathematical Competitions

From the A–H inequality we get
$$\begin{aligned} \frac{a+1}{a(a+2)} + \frac{b+1}{b(b+2)} + \frac{c+1}{c(c+2)} &\ge \frac{9}{\frac{a(a+2)}{a+1} + \frac{b(b+2)}{b+1} + \frac{c(c+2)}{c+1}} \\ &= \frac{9}{\frac{(a+1)^2-1}{a+1} + \frac{(b+1)^2-1}{b+1} + \frac{(c+1)^2-1}{c+1}} \\ &= \frac{9}{(a+1) + (b+1) + (c+1) - \left(\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}\right)} \end{aligned}$$
Since $(a+1) + (b+1) + (c+1) = a+b+c+3 \le 6$ and
$$\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} \ge \frac{9}{(a+1) + (b+1) + (c+1)} \ge \frac{9}{6} = \frac{3}{2}$$
we can conclude
$$\frac{9}{(a+1) + (b+1) + (c+1) - \left(\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}\right)} \ge \frac{9}{6 - \frac{3}{2}} = 2$$
which finishes the proof.