Proving an inequality using am-gm inequality

Question

Prove that if $x\neq y\neq z$, $x,y,z\gt0$ $$\frac{x^2y^2+y^2z^2+z^2y^2}{x+y+z}\geqslant xyz$$

I tried using $$x^2y^2+y^2z^2+z^2x^2\gt 3\root3\of{x^4y^4z^4}$$ $$x+y+z\gt 3\root3\of{xyz}$$ But I can't divide both LHS and RHS

Then I tried to prove that $x^2y^2+y^2z^2+z^2x^2\geqslant xyz(x+y+z)$, so I used $$x^2y^2+y^2z^2+z^2x^2\gt 3\root3\of{x^4y^4z^4}$$ $$xyz(x+y+z) \gt 3\root3\of{x^4y^4z^4}$$

Then I got stuck as I can't subtract the two inequalities either (only addition or multiplication)

Answer 1

Using the known inequality $$a^2+b^2+c^2\geq ab+bc+ca$$ we get $$(xy)^2+(yz)^2+(zx)^2\geq xyyz+xyxz+yzzx=xy^2z+x^2yz+xyz^2=(xyz)(x+y+z)$$ The first inequality is equivalent to $$(a-b)^2+(b-c)^2+(c-a)^2\geq 0$$

Answer 2

By AM-GM $$\sum_{cyc }x^2y^2=\frac{1}{2}\sum_{cyc}(x^2y^2+x^2z^2)\geq\frac{1}{2}\sum_{cyc}2\sqrt{x^2y^2\cdot x^2z^2}=\sum_{cyc}x^2yz=xyz(x+y+z).$$