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Ukrainian National Mathematical Olympiad, Third Round, First Tour

Algebra Difficulty 5.1 AIME, harder Prove it Ukraine

For positive real numbers $a$ , $b$ , $c$ , which satisfy the condition $ab + bc + ca = 1$ , prove the following inequality:
$\left(\sqrt{bc} + \frac{1}{2a + \sqrt{bc}}\right) \cdot \left(\sqrt{ca} + \frac{1}{2b + \sqrt{ca}}\right) \cdot \left(\sqrt{ab} + \frac{1}{2c + \sqrt{ab}}\right) \ge 8abc.$

Solution

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.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.