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. $2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x)$
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. $2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x)$
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 $x^4 + y^4 + z^4 \ge x^3y+y^3z+z^3x$.
If you set $x=0$ and go from there ... [always test some easy values first]
Define $f(x,y,z)=x^2y^2+y^2z^2+z^2x^2-x^3y-y^3z-z^3x$. Then $f(1,1,2)=-2$ and $f(1,1,-1)=4$. Therefore, neither the inequality nor its reverse is true.
For positive numbers, we also have $f(3,2,1)=-16$ and $f(4,6,1)=24$. So even for all positive numbers the inequality is not true.