<s>[INST] Prove/disprove $\sum x^2y^2 \ge \sum x^3y$ Prove $x^2y^2+y^2z^2+z^2x^2 \ge$  or $\le x^3y+y^3z+z^3x$ where $x,y,z$ are real numbers.
Actually, I have reached here from this problem:
Inequality. $2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x)$
 [/INST] If you set x = 0 and y = 1, then $x^2y^2+y^2z^2+z^2x^2$ becomes $z^2$ and $x^3y+y^3z+z^3x$ becomes $z$. Clearly the relation between them depends on the value of $z$. As for the original inequality, you can prove that $x^4 + y^4 + z^4 +x^2y^2+y^2z^2+z^2x^2\ge 2(x^3y+y^3z+z^3x)$ and $x^4 + y^4 + z^4 \ge x^3y+y^3z+z^3x$.
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