Maths Olympiad Prep

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Algebra Difficulty 6.1 National olympiad Prove it

Example 2.1.10. Let a,b,ca, b, c be the side-lengths of a triangle. Prove that
a3a−b+c+b3b−c+a+c3c−a+b≥1. \frac{a}{3a - b + c} + \frac{b}{3b - c + a} + \frac{c}{3c - a + b} \geq 1.

Solution

Solution. By Cauchy-Schwarz, we have
4∑cyca3a−b+c=∑cyc4a3a−b+c=3+∑cyca+b−c3a−b+c≥3+(a+b+c)2∑cyc(a+b−c)(3a−b+c)=4\begin{aligned} 4 \sum_{c y c} \frac{a}{3 a-b+c} & =\sum_{c y c} \frac{4 a}{3 a-b+c} \\ & =3+\sum_{c y c} \frac{a+b-c}{3 a-b+c} \\ & \geq 3+\frac{(a+b+c)^{2}}{\sum_{c y c}(a+b-c)(3 a-b+c)} \\ & =4 \end{aligned}

Equality holds for a=b=ca=b=c.

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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.