First of all,

$\displaystyle\begin{align}\sum_{cycl}c\left(\frac{4a}{b^2}+\frac{3b}{a^2}\right) &= \sum_{cycl}\frac{4ac}{b^2} + \sum_{cycl}\frac{3bc}{a^2}\\ &=4\sum_{cycl}\frac{ab}{c^2} + 3\sum_{cycl}\frac{ab}{c^2}\\ &=7\sum_{cycl}\frac{ab}{c^2}. \end{align}$

By the AM-GM inequality,

$\displaystyle\begin{align} 4\sum_{cycl}\frac{ab}{c^2}&\ge 4\cdot 3\sqrt[3]{\frac{ab}{c^2}\cdot \frac{bc}{a^2}\cdot \frac{ca}{b^2}}\\ &=12\sqrt[3]{1}\\ &=12. \end{align}$

On the other hand,

$\displaystyle\begin{align} \frac{ab}{c^2}+\frac{ab}{c^2}+\frac{ca}{b^2}&\ge 3\sqrt[3]{\frac{a^3}{c^3}}=3\frac{a}{c},\\ \frac{bc}{a^2}+\frac{bc}{a^2}+\frac{ab}{c^2}&\ge 3\sqrt[3]{\frac{b^3}{a^3}}=3\frac{b}{a},\\ \frac{ca}{b^2}+\frac{ca}{b^2}+\frac{bc}{a^2}&\ge 3\sqrt[3]{\frac{c^3}{b^3}}=3\frac{c}{b},\\ \end{align}$

Adding these to (1) we get

$\displaystyle\begin{align} 7\sum_{cycl}\frac{ab}{c^2} &= 4\sum_{cycl}\frac{ab}{c^2} + 3\sum_{cycl}\frac{ab}{c^2}\\ &\ge 12 + 3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right). \end{align}$

The problem above has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru (from his book Math Accent), along with a solution by Soumava Chakraborty.