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. $2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x)$
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I think the reverse inequality is actually true, and be carefull your inequality is not equivalent to the original one. Disproving this means nothing for that one.... – N. S. Sep 25 '12 at 18:43
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@N.S., I have edited the question, how to prove the reverse inequality. – lab bhattacharjee Sep 25 '12 at 18:46
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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.
robjohn
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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 $x^4 + y^4 + z^4 \ge x^3y+y^3z+z^3x$.
corindo
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ya, using AM, GM. But these 2 inequalities does not help us to reach at the problem specified, right? – lab bhattacharjee Sep 25 '12 at 18:49
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Eh....Add up these 2 inequalities and add $3(x^2y^2 + y^2z^2 + z^2x^2)$ to both sides. And you will find your inequality. – corindo Sep 25 '12 at 18:54