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Let $x,y,z\in \mathbb R^+$ prove that:

$$\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge xyz + \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$

The inequality $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \overset{C-S}{\ge} 3xyz$ will not help because $3xyz \le RHS$. How to prove above inequality ? It is too hard for me.

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The inequality is homogeneous, so we may assume that $xyz=1$. Then if we expand left and right hand side, we see that we have to show that $$\sqrt{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 3} \ge 1 + \sqrt[3]{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 2}.$$ Let $$u = \sqrt[3]{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 2}.$$ If we square both sides of the expanded inequality, we get the equivalent inequality $$u^3 \ge 2u + u^2,$$ or, $$u^2 - u - 2 \ge 0.$$ Therefore it is enough to show that $u \ge 2$. But this is certainly true since $x^3 + \frac{1}{x^3}, y^3 + \frac{1}{y^3}, z^3 + \frac{1}{z^3} \ge 2$.

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  • $\begingroup$ $x=y=z=2$ violates your assumption, no? $\endgroup$ – response May 30 '13 at 15:41
  • $\begingroup$ @response what do you mean? I'm allowed to do that assumption because I'm allowed to multiply all of the variables $x,y,z$ by a constant of my choicing (because of the form of the inequality). In particular I'm allowed to divide all of them by $2$ before I start. $\endgroup$ – J. J. May 30 '13 at 15:44
  • $\begingroup$ @response What he means is the following: let $x'=\frac{x}{\sqrt[3]{xyz}}$ and so on. Then the inequality for $x',y',z'$ is exactly the same.... $\endgroup$ – N. S. May 30 '13 at 15:44
  • $\begingroup$ I see it now. Somehow I glossed over the part where you talk about it being a 'homogeneous' system. $\endgroup$ – response May 30 '13 at 15:45
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$$\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \overset{AM-GM}{\ge} 3xyz$$

DOES help, since you can also prove:

$$2\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \geq 3 \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$

Note that this is the same as

$$2^6(x^2y+ y^2z+z^2x)^3(y^2x+z^2y+x^2z)^3 \geq 3^6x^2y^2z^2(x^2+yz)^2(y^2+xz)^2(z^2+xy)^2 \,.$$

which after long computations and many tears can be reduced to this all junk being positive. If you look to the few negative terms , you can get easely rid of them by invoking the AM-GM inequality.

Now, all this suggests that the inequality

$$2\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \geq 3 \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$

is true, all you have to do is find a neater solution :)

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  • $\begingroup$ heh, +1 for the work :) $\endgroup$ – J. J. May 30 '13 at 15:58
  • $\begingroup$ @J.J. You should give a +1 to WA then :D $\endgroup$ – N. S. May 30 '13 at 15:59
  • $\begingroup$ I assume you at least checked that you really can bunch the terms. :) $\endgroup$ – J. J. May 30 '13 at 16:00
  • $\begingroup$ @J.J. Yea, but since the first alternate form is a product of huge polynomials, and I only saw 1 minus, in only took me few seconds.... $\endgroup$ – N. S. May 30 '13 at 16:01
  • $\begingroup$ Oh, you're right indeed. I stared a while at the second form. :P $\endgroup$ – J. J. May 30 '13 at 16:02

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