# A Cyclic Inequality in Three Variables with Constraint

### Problem ### Solution

In any triangle $ABC,\,$ $A+B+C=\pi\,$ and

$\displaystyle\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1.$

Let's find $A,B,C\,$ such that $a\sqrt{bc}=\cos^2A,\,$ $b\sqrt{ca}=\cos^2B,\,$ $c\sqrt{ab}=\cos^2C.\,$ Assuming this done, $abc=\cos A\cos B\cos C\gt 0\,$ (so that, if $\Delta ABC\,$ exists, it is acute.) We can then find

\displaystyle\begin{align} a &=\frac{\cos^3 A}{\cos B\cos C}\gt 0,\\ b &=\frac{\cos^3 B}{\cos C\cos A}\gt 0,\\ c &=\frac{\cos^3 C}{\cos A\cos B}\gt 0 \end{align}

It is directly verified that these $a,b,c\,$ satisfy the constraint. The required inequality reduces to

$\displaystyle\prod_{cycl}\frac{\cos^2B\cos^2C}{\cos^2A}\ge 2\prod_{cycl}\frac{\cos^3A}{\cos B\cos C}.$

In any $\Delta ABC,\,$ with positive $x,y,z,\,$

$x^2+y^2+z^2\ge 2xy\cos C+2yz\cos A+2zx\cos B.$

Taking, $x=\displaystyle\frac{\cos B\cos C}{\cos^2A},\,$ $y=\displaystyle\frac{\cos C\cos A}{\cos^2B},\,$ $z=\displaystyle\frac{\cos A\cos B}{\cos^2C},\,$ we exactly get

$\displaystyle\sum_{cycl}\frac{\cos^2B\cos^2C}{\cos^4A}\ge 2\sum_{cycl}\frac{\cos B\cos C}{\cos^2A}\frac{\cos C\cos A}{\cos^2B}\cos C=2\sum_{cycl}\frac{\cos^3C}{\cos A\cos B}.$

### Acknowledgment

Dan Sitaru has kindly posted the above problem (from the Romanian Mathematical Magazine, #JP033) at the CutTheKnotMath facebook page. The problem is due to Nguyen Viet Hung; Solution is by Kevin Soto Palacios. Copyright © 1996-2018 Alexander Bogomolny

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