# Dorin Marghidanu's Cyclic Inequality in Three Variables III

### Solution 1

\displaystyle\begin{align} \sum_{cycl}\frac{a^3}{b^2+c} &= \sum_{cycl}\frac{a^4}{ab^2+ac}& \text{by Bergstrom's inequality}\\ &\ge\frac{(a^2+b^2+c^2)^2}{ab^2+bc^2+ca^2+ab+bc+ca}&\text{since }a,b,c\in (0,1)\\ &\ge\frac{(a^2+b^2+c^2)^2}{2(ab+bc+ca)}\\ &\ge\frac{(a^2+b^2+c^2)^2}{2(a^2+b^2+c^2)}\\ &=\frac{1}{2}(a^2+b^2+c^2)\ge\frac{3}{2}\sqrt[3]{a^2b^2c^2}.&\text{by the AM-GM inequality}\\ \end{align}

### Solution 2

$a,b,c\in (0,1)$ implies $a^2\lt a,$ $b^2\lt b,$ $c^2\lt c.$ WLOG, $a\ge b\ge c,$ so that

\displaystyle \begin{align} &a^2\ge b^2\ge c^2\\ &a+b\ge a+c\ge b+c\\ &\frac{1}{b+c}\ge \frac{1}{c+a}\ge\frac{1}{a+b}\\ &\frac{a}{b+c}\ge \frac{b}{c+a}\ge\frac{c}{a+b}. \end{align}

It follows that

\displaystyle \begin{align} \sum_{cycl}\frac{a^3}{b^2+c}&\gt\sum_{cycl}\frac{a\cdot a^2}{b+c}&\text{by Chebyshev's inequality}\\ &\ge\frac{\displaystyle \sum_{cycl}a^2}{3}\cdot\sum_{cycl}\frac{a}{b+c}&\text{AM-GM + Nesbitt}\\ &\ge\frac{3(abc)^{\frac{2}{3}}}{3}\cdot\frac{3}{2}\\ &=\frac{3}{2}(abc)^{\frac{2}{3}}. \end{align}

### Solution 3

From $a,b,c\in (0,1),$ $a^2\lt a,$ $b^2\lt b,$ $c^2\lt c.$ With these and, successively, the generalized Radon's an the AM-GM inequalities,

\displaystyle \begin{align} \sum_{cycl}\frac{a^3}{b^2+c}&\gt\sum_{cycl}\frac{a^3}{b+c}\\ &\ge\frac{1}{3}\cdot\frac{(a+b+c)^3}{2(a+b+c)}=\frac{1}{6}(a+b+c)^2\\ &\ge\frac{1}{6}\cdot 3^2(abc)^{\frac{2}{3}}=\frac{3}{2}(abc)^{\frac{2}{3}}. \end{align}

### Acknowledgment

This problem has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu. Solution 2 is by Uche E. Okeke, Solution 3 is by Dorin Marghidanu.