62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour

From the statement it follows, that $$\frac{1}{2a+\sqrt{bc}}=\frac{ab+bc+ca}{2a+\sqrt{bc}}=\frac{bc+a(b+c)}{2a+\sqrt{bc}}\ge \frac{bc+2a\sqrt{bc}}{2a+\sqrt{bc}}=\sqrt{bc},$$ implying $$\sqrt{bc}+\frac{1}{2a+\sqrt{bc}}\ge 2\sqrt{bc}.$$ If we write two similar inequalities $$\sqrt{ca}+\frac{1}{2b+\sqrt{ca}}\ge 2\sqrt{ca},$$ $$\sqrt{ab}+\frac{1}{2c+\sqrt{ab}}\ge 2\sqrt{ab},$$ and multiply them, we get the desired inequality.