Prove that $xy+yz+zx+\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\ge 2x^2+2y^2+2z^2$ for every $x,y,z$ strictly positive I checked the inequality for $y=z$. In this particular case, after some simplifications, the inequality becomes:
$$
2x^3+y^3\ge 3x^2y,
$$
which is true, according to the arithmetic-geometric mean inequality applied to the numbers $x^3,\ x^3$ and $y^3$. I have no idea, at least for now, on how to proceed in the general case. Please give me a hint.
 A: Let $\frac{xy}{z}=a^2$, $\frac{yz}{x}=b^2$, and $\frac{zx}{y}=c^2$.
Now, we have $y=ab$, $z=bc$, and $x=ca$. 
It now suffices to prove $$a^4+b^4+c^4+a^2bc+b^2ca+c^2ab \ge 2(a^2b^2+b^2c^2+c^2a^2)$$ 
From Schur's Inequality, we have $$a^2(a-b)(a-c)+b^2(b-c)(b-a)+c^2(c-a)(c-b)=(a^4+b^4+c^4)-(a^3b+a^3c+b^3c+b^3a+c^3a+c^3b)+(a^2bc+b^2ca+c^2ab) \ge 0$$
This gives us $$a^4+b^4+c^4+a^2bc+b^2ca+c^2ab \ge (a^3b+b^3a)+(b^3c+c^3b)+(c^3a+a^3c) \ge 2(a^2b^2+b^2c^2+c^2a^2)$$
as desired.
A: Use Schur inequality
$$a^4+b^4+c^4+abc(a+b+c)\ge \sum_{cyc}(a^3b+ab^3)\ge \sum_{cyc}2a^2b^2$$
Let
$$a=xy,b=yz,c=xz$$
then is your inequality
A: simplifying and factorizing the term $$xy+yz+zx+\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}-2(x^2+y^2+z^2)$$ we get
$${\frac { \left( xy+zx+yz \right)  \left( {y}^{3}{x}^{3}-{y}^{2}z{x}^{3
}-y{z}^{2}{x}^{3}+{z}^{3}{x}^{3}-{y}^{3}z{x}^{2}+3\,{z}^{2}{x}^{2}{y}^
{2}-y{z}^{3}{x}^{2}-{y}^{3}{z}^{2}x-{z}^{3}{y}^{2}x+{y}^{3}{z}^{3}
 \right) }{{z}^{2}{x}^{2}{y}^{2}}}$$
it must be
$${y}^{3}{x}^{3}-{y}^{2}z{x}^{3}-y{z}^{2}{x}^{3}+{z}^{3}{x}^{3}-{y}^{3}z
{x}^{2}+3\,{z}^{2}{x}^{2}{y}^{2}-y{z}^{3}{x}^{2}-{y}^{3}{z}^{2}x-{z}^{
3}{y}^{2}x+{y}^{3}{z}^{3}\geq 0$$
assume that $x=\min(x,y,z)$ and set $y=x+u,z=x+y+z$ after simplification we obtain
$$\left( {u}^{2}+uv+{v}^{2} \right) {x}^{4}+ \left( 4\,{u}^{3}+6\,{u}^{
2}v+2\,u{v}^{2} \right) {x}^{3}+ \left( 6\,{u}^{4}+12\,{u}^{3}v+6\,{u}
^{2}{v}^{2} \right) {x}^{2}+ \left( 4\,{u}^{5}+10\,{u}^{4}v+8\,{u}^{3}
{v}^{2}+2\,{u}^{2}{v}^{3} \right) x+{u}^{6}+3\,{u}^{5}v+3\,{u}^{4}{v}^
{2}+{u}^{3}{v}^{3}
\geq 0$$ since $u,v\geq 0$.
