Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.8 AIME, harder Prove it

Through a point lying inside a triangle, three lines parallel to its sides are drawn. Let's denote the areas of the parts into which these lines divide the triangle as shown in the figure. Prove that $a / \boldsymbol{\alpha} +$ $b / \beta + c / \gamma \geq 3 / 2$

Solution

According to the inequality between the geometric mean and the arithmetic mean $\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma} \geq 33 a b c /(\boldsymbol{\alpha} \boldsymbol{\beta} \gamma$ ) = 3/2, since $\alpha=2 \sqrt{b c}, \beta=2 \sqrt{c a}$ and $\gamma=2 \sqrt{a b}$ (see problem $\underline{1.33}$ ).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.