Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Prove it Ukraine

For positive real numbers aa, bb, cc, which satisfy the condition ab+bc+ca=1ab + bc + ca = 1, prove the following inequality:
(bc+12a+bc)⋅(ca+12b+ca)⋅(ab+12c+ab)≥8abc. \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
12a+bc=ab+bc+ca2a+bc=bc+a(b+c)2a+bc≥bc+2abc2a+bc=bc, \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
bc+12a+bc≥2bc. \sqrt{bc} + \frac{1}{2a + \sqrt{bc}} \ge 2\sqrt{bc}.
If we write two similar inequalities
ca+12b+ca≥2ca, \sqrt{ca} + \frac{1}{2b + \sqrt{ca}} \ge 2\sqrt{ca},
ab+12c+ab≥2ab, \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.