Croatian Mathematical Olympiad

The inequality is homogeneous, hence without loss of generality, we can assume that $a+b+c=1$. The given inequality transforms into
$$\underbrace{\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}}_{A} + \underbrace{\frac{\sqrt{a}}{1-a} + \frac{\sqrt{b}}{1-b} + \frac{\sqrt{c}}{1-c}}_{B} \ge \frac{9+3\sqrt{3}}{2}.$$
By the inequality of arithmetic and harmonic means, we have
$$A \ge \frac{9}{3 - (a + b + c)} = \frac{9}{2}.$$
On the other hand, by the Cauchy-Bunyakovsky-Schwarz inequality, we have
$$B \cdot \left( a\sqrt{a}(1-a) + b\sqrt{b}(1-b) + c\sqrt{c}(1-c) \right) \ge 1,$$
and it suffices to show that
$$a\sqrt{a}(1-a) + b\sqrt{b}(1-b) + c\sqrt{c}(1-c) \le \frac{2}{3\sqrt{3}}.$$
By the inequality of geometric and arithmetic means, we have
$$\frac{a(1-a)^2}{4} \le \left( \frac{a + \frac{1-a}{2} + \frac{1-a}{2}}{3} \right)^3 = \frac{1}{27},$$
i.e. $\sqrt{a}(1-a) \le \frac{2}{3\sqrt{3}}$, and analogously $\sqrt{b}(1-b) \le \frac{2}{3\sqrt{3}}$, $\sqrt{c}(1-c) \le \frac{2}{3\sqrt{3}}$.
Finally,
$$a\sqrt{a}(1-a) + b\sqrt{b}(1-b) + c\sqrt{c}(1-c) \le (a+b+c) \cdot \frac{2}{3\sqrt{3}} = \frac{2}{3\sqrt{3}},$$
and the given inequality is proved.