Prove $\sum\limits_{cyc} \frac{(a+14)^2}{(b+4)^2}\geq \frac{3}{2} \sum\limits_{cyc} \left(\frac{a+14}{b+4}+\frac{b+4}{a+14}\right)-6$

Question

Problem. For $a,b,c\geq 0.$ Prove that: $$\sum\limits_{cyc} \frac{(a+14)^2}{(b+4)^2}\geq \frac{3}{2} \sum\limits_{cyc} \left(\frac{a+14}{b+4}+\frac{b+4}{a+14}\right)-6$$ I have a solution but it's not really nice. The main idea is take $\frac{a+14}{b+4}=x,..$ and try to use the familiar result by Vo Quoc Ba Can.

You can find it in my blog. I'm looking for alternative solution now.

Thanks a lot!