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 $xyz=1$

Question

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

Answer 1

By arithmetic mean-geometric mean inequality: $y^2z^2 + z^2x^2 \geq 2z^2xy = 2z$. Also, we can use Sedrakyan's Form of Cauchy-Schwarz inequality and arithmetic mean-geometric mean inequality we get

$S= \cfrac{1}{(y+z) x^4} + \cfrac{1}{(x+z) y^4} + \cfrac{1}{(y+x) z^4} = \cfrac{(xyz)^4}{(y+z) x^4} + \cfrac{(xyz)^4}{(x+z) y^4} + \cfrac{(xyz)^4}{(y+x) z^4}\\ = \cfrac{(yz)^4}{(y+z)} + \cfrac{(xz)^4}{(x+z)} + \cfrac{(xy)^4}{(y+x)} \geq \dfrac{(y^2z^2 + z^2x^2 + x^2y^2)^2}{2(x+y+z)} \geq \dfrac{(x+y+z)^2}{2(x+y+z)} = \dfrac{1}{2}(x+y+z) \geq \dfrac{3}{2}\sqrt[3]{xyz} = \dfrac{3}{2}.$

Equality holds when $x=y=z=1$.

Answer 2

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