How do I prove that :

$$2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x).$$

It seems to be easy , but I have no idea:)

thanks :)


2 Answers 2


I assume that $x,y,z>0$, even if not explicit from your post. Tell me if I'm wrong.

Expand the $\operatorname{LHS}$. Then we want to prove that

$$2x^4+2y^4+2z^4+4x^2y^2+4y^2z^2+4z^2x^2\geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x).$$

It is sufficient to estabilish

$$2x^4+2y^4+2z^4+x^2y^2+y^2z^2+z^2x^2\geq 3(x^3y+y^3z+z^3x).$$

Since $$\begin{split}x^4+x^2y^2&\geq 2x^3y,\\y^4+y^2z^2&\geq 2y^3z,\\z^4+z^2x^2&\geq 2z^3x,\end{split}$$ by the arithmetic-geometric inequality, we are reduced to prove that

$$x^4+y^4+z^4\geq x^3y+y^3z+z^3x.$$

This follows from Muirhead inequality (bunching), however we can give a direct proof again by AM-GM

$$\begin{split}&\frac{x^4}{4}+\frac{x^4}{4}+\frac{x^4}{4}+\frac{y^4}{4}\geq x^3y\\&\frac{y^4}{4}+\frac{y^4}{4}+\frac{y^4}{4}+\frac{z^4}{4}\geq y^3z\\\\&\frac{z^4}{4}+\frac{z^4}{4}+\frac{z^4}{4}+\frac{x^4}{4}\geq z^3x.\end{split}$$

Sum up to conclude.

  • $\begingroup$ @labbhattacharjee You gave a nice solution, why did you delete it ? My friend wants your solution, it is one nice. can you post again the solution? thanks:) $\endgroup$
    – Iuli
    Commented Sep 25, 2012 at 16:17
  • $\begingroup$ @Iuli, though it seemed nice, it was not error-free. $\endgroup$ Commented Sep 25, 2012 at 16:21
  • $\begingroup$ nice! I think it is not necessary for x,y,z>0,as each step shows even x,y,z are not positive,the inequality is OK. $\endgroup$
    – chenbai
    Commented Apr 6, 2013 at 7:08

We start the same way as uforoboa: Expanding the left hand side, you are trying to show that $$2x^4 + 2y^4 + 2z^4 + 4(x^2y^2+y^2z^2+z^2x^2) \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x)$$ This equivalent to $$2x^4 + 2y^4 + 2z^4 +x^2y^2+y^2z^2+z^2x^2 \geq 3(x^3y+y^3z+z^3x)$$ By AM-GM one has $x^3y \leq 1/2(x^4 + x^2y^2)$, $y^3z \leq 1/2(y^4 + y^2z^2)$, and $z^3x \leq 1/2(z^4 + z^2x^2)$. Inserting this into the above it suffices to show that $$2x^4 + 2y^4 + 2z^4 +x^2y^2+y^2z^2+z^2x^2 \geq {3 \over 2}x^4 + {3 \over 2}y^4 + {3 \over 2}z^4 +{3 \over 2}x^2y^2+{3 \over 2}y^2z^2+{3 \over 2}z^2x^2$$ This is the same as showing $$x^4 + y^4 + z^4 \geq x^2y^2+y^2z^2+z^2x^2$$ This follows from AM-GM on each term on the right, or just expanding $(x^2 - y^2)^2 + (y^2 - z^2)^2 + (x^2 - z^2)^2 \geq 0$.


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