For $x,y,z$. Prove that $\frac{1}{x^2+4yz}+\frac{1}{y^2+4zx}+\frac{1}{z^2+4xy}\le \frac{1}{xyz}$

Question

For $x,y,z$ are postive real number satisfy $x+y+z=4$. Prove that $$\frac{1}{x^2+4yz}+\frac{1}{y^2+4zx}+\frac{1}{z^2+4xy}\le \frac{1}{xyz}$$

Answer 1

by AM-GM we have $$\frac{x^2+4yz}{2}\geq 2x\sqrt{yz}$$ and so on, thus we have $$\frac{1}{x^2+4yz}+\frac{1}{y^2+4zx}+\frac{1}{z^2+4xy}\le \frac{1}{4}\left(\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\right)$$ we have to Show that this is $$\le 1$$ by $AM-GM$ we have again $$\frac{1}{4}\left(\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\right)\le \frac{1}{4}(x+y+z)=1$$ since $$x+y+z=4$$ i want to insert one line from $$\frac{1}{4x\sqrt{yz}}+\frac{1}{4y\sqrt{xz}}+\frac{1}{4z\sqrt{xy}}\le \frac{1}{xyz}$$ follows the inequauality (after multiplication by $$xyz$$) $$\frac{1}{4}\left(\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\right)\le 1$$