# Cyclic Inequality with Arctangents

### Solution

Let $f(x)=\arctan(x).$ Then $\displaystyle f'(x)=\frac{1}{1+x^2}\gt 0$ and $\displaystyle f''(x)=-\frac{2x}{1+x^2}\le 0,$ for $x\ge 0.$ Thus, for $x\ge 0,$ the function is increasing and concave. We are in a position to apply Jensen's inequality:

\displaystyle \begin{align} \sum_{cycl}\frac{ab}{ab+bc+ca}\cdot\arctan\frac{c}{b}&\le\arctan\left(\sum_{cycl}\frac{ab}{ab+bc+ca}\cdot\frac{c}{b}\right)\\ &=\arctan\left(\sum_{cycl}\frac{ac}{ab+bc+ca}\right)\\ &=\arctan(1)=\frac{\pi}{4} \end{align}

Further, by the Rearrangement inequality, in particular,

$\displaystyle 4\sum_{cycl}ab\cdot\arctan\frac{c}{b}\le\pi (ab+bc+ca)\le \pi(a^2+b^2+c^2).$

Equality is attained for $a=b=c.$

### Acknowledgment

This is a slightly modified version of Dan Sitaru's inequality that kindly posted at the CutTheKnotMath facebook page. The original inequality was previously published at the Romanian Mathematical Magazine. The above solution is by Diego Alvariz.