How do I show that $\frac {\cos^2 A}{\cos^2 B} + \frac {\cos^2 B}{\cos^2 C} + \frac {\cos^2 C}{\cos^2 A} \ge 4(\cos^2 A + \cos^2 B + \cos^2 C)$?

Question

Let $A, B, C$ be the angles of an acute triangle. Show that $$\frac {\cos^2 A}{\cos^2 B} + \frac {\cos^2 B}{\cos^2 C} + \frac {\cos^2 C}{\cos^2 A} \ge 4(\cos^2 A + \cos^2 B + \cos^2 C).$$

How should I approach this kind of "geometric inequalities"? I've considered substituting the cosines using the law of the cosines and then use the Ravi transformation to turn it into an algebraic one, but that seems too tedious and unlikely to yield any beautiful solution. Any hints will be appreciated!

Answer 1

By AM-GM easy to show that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq\frac{x+y+z}{\sqrt[3]{xyz}}$ for all positives $x$, $y$ and $z$ and

$\cos\alpha\cos\beta\cos\gamma\leq\frac{1}{8}$.

Hence, $\sum\limits_{cyc}\frac{\cos^2\alpha}{\cos^2\beta}\geq\frac{\cos^2\alpha+\cos^2\beta+\cos^2\gamma}{\sqrt[3]{\cos^2\alpha\cos^2\beta\cos^2\gamma}}\geq4\left(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\right)$. Done!