Iranian Mathematical Olympiad

$$\frac{1}{2a+\sqrt{bc}}=\frac{ab+bc+ca}{2a+\sqrt{bc}}.$$ Due to LM-GM, the inequality $$ab+ca=a(b+c)\ge a\times 2\sqrt{bc}$$ holds. Hence $$\frac{1}{2a+\sqrt{bc}}=\frac{ab+bc+ca}{2a+\sqrt{bc}}\ge \frac{2a\sqrt{bc}+bc}{2a+\sqrt{bc}}=\sqrt{bc},$$ and so $$\sqrt{bc}+\frac{1}{2a+\sqrt{bc}}\ge 2\sqrt{bc}.$$ Similarly, it is deduced that $$\sqrt{ca}+\frac{1}{2b+\sqrt{ca}}\ge 2\sqrt{ca},$$ and $$\sqrt{ab}+\frac{1}{2c+\sqrt{ab}}\ge 2\sqrt{ab}.$$ Multiplying the latest three inequalities leads to the desired conclusion. ■