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 $$\underset{A}{\underbrace{\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}}}+\underset{B}{\underbrace{\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}}}\ge \frac{9+3\sqrt{3}}{2}.$$ 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.