Prove $\frac{1}{(y+z) x^4} + \frac{1}{(x+z) y^4} + \frac{1}{(y+x) z^4}\geq\frac{3}{2}$ for $x, y, z>0$, with $xyz=1$ How to prove that $\frac{1}{(y+z) x^4} + \cfrac{1}{(x+z) y^4} + \cfrac{1}{(y+x) z^4}\geq\frac{3}{2}$ for $x, y, z>0$...
I've tried substituting $a=1/x$, $b=1/y,$ $c=1/z$, and also writing the 1 that is being divided into $(xyz)^4.$ In both of those cases I tried applying Titu's lemma, but didn't manage to do anything. In the first case, I've observed that it is similar to Nesbitt's inequality, just with an extra $a^2$,$b^2$ and $c^2$ multiplied

By Holder and AM-GM we obtain:
$$\sum_{cyc}\frac{1}{(y+z)x^4}=\frac{\sum\limits_{cyc}\frac{1}{(y+z)x^4}\sum\limits_{cyc}(y+z)x\sum\limits_{cyc}1}{6(xy+xz+yz)}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}\frac{1}{x}\right)^3}{6(xy+xz+yz)}=\frac{(xy+xz+yz)^2}{6}\geq\frac{9}{6}=\frac{3}{2}.$$
