<s>[INST] 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.
 [/INST] 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.
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