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Taiwan IMO Selection Camp

Algebra Difficulty 4.6 AIME Prove it Taiwan

Given $a, b, c, d > 0$ , prove that:
$\sum_{cyc} \frac{c}{a + 2b} + \sum_{cyc} \frac{a + 2b}{c} \geq 8 \left( \frac{(a + b + c + d)^2}{ab + ac + ad + bc + bd + cd} - 1 \right),$
where $\sum_{cyc} f(a, b, c, d) = f(a, b, c, d) + f(d, a, b, c) + f(c, d, a, b) + f(b, c, d, a)$

Solution

Notice that
$\frac{c}{a+2b} = \frac{a+2b+c}{a+2b} - 1, \quad \frac{a+2b}{c} = \frac{a+2b+c}{c} - 1$
So we have:
$\begin{aligned} \sum_{cyc} \frac{c}{a+2b} + \sum_{cyc} \frac{a+2b}{c} &= \sum_{cyc} (a+2b+c) \left( \frac{1}{a+2b} + \frac{1}{c} \right) - 8 \\ &= \sum_{cyc} \frac{(a+2b+c)^2}{c(a+2b)} - 8. \end{aligned}$
Using the Cauchy-Schwarz inequality:
$\left( \sum_{cyc} c(a+2b) \right) \left( \sum_{cyc} \frac{(a+2b+c)^2}{c(a+2b)} \right) \geq 16(a+b+c+d)^2$
that is
$\begin{aligned} \sum_{cyc} \frac{(a+2b+c)^2}{c(a+2b)} - 8 &\geq 8 \left( \frac{(a+b+c+d)^2}{\sum_{cyc} c(a+2b)} - 1 \right) \\ &= 8 \left( \frac{(a+b+c+d)^2}{ab+ac+ad+bc+bd+cd} - 1 \right). \end{aligned}$

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from the original; metadata (topic, difficulty) added by this project.