For $x,y,z > 0$ and $xyz=1$, prove that \[\frac{x^{9}+y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}+\frac{x^{9}+z^{9}}{x^{6}+x^{3}z^{3}+z^{6}}+\frac{y^{9}+z^{9}}{y^{6}+y^{3}z^{3}+z^{6}}\geq 2\]
Problem
Source: Romania
Tags: inequalities
SplashD
05.11.2006 19:53
Jan wrote: For $x,y,z > 0$ and $xyz=1$, prove that \[\frac{x^{9}+y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}+\frac{x^{9}+z^{9}}{x^{6}+x^{3}z^{3}+z^{6}}+\frac{y^{9}+z^{9}}{y^{6}+y^{3}z^{3}+z^{6}}\geq 2 \]
$\Leftrightarrow \frac{(x^{9}-y^{9})+2y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}+\frac{(z^{9}-x^{9})+2x^{9}}{x^{6}+x^{3}z^{3}+z^{6}}+\frac{(y^{9}-z^{9})+2z^{9}}{y^{6}+y^{3}z^{3}+z^{6}}\geq 2$
$\Leftrightarrow \frac{2y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}+\frac{2x^{9}}{x^{6}+x^{3}z^{3}+z^{6}}+\frac{2z^{9}}{y^{6}+y^{3}z^{3}+z^{6}}\geq 2$
$\Leftrightarrow \frac{y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}+\frac{x^{9}}{x^{6}+x^{3}z^{3}+z^{6}}+\frac{z^{9}}{y^{6}+y^{3}z^{3}+z^{6}}\geq xyz$
But $\frac{y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}+\frac{x^{9}}{x^{6}+x^{3}z^{3}+z^{6}}+\frac{z^{9}}{y^{6}+y^{3}z^{3}+z^{6}}\ge \frac{(x^{6}+y^{6}+z^{6})^{2}}{\sum_{cyc}(x^{9}+x^{6}y^{3}+x^{3}y^{6})}$ by Cauchy
So it remains to prove:
$(x^{6}+y^{6}+z^{6})^{2}\ge xyz \sum_{cyc}(x^{9}+x^{6}y^{3}+x^{3}y^{6})$
$\Leftrightarrow x^{12}+y^{12}+z^{12}+2x^{6}y^{6}+2y^{6}z^{6}+2z^{6}x^{6}\ge \sum_{cyc}x^{10}yz+\sum_{sym}x^{7}y^{4}z$
$\Leftrightarrow \sum_{cyc}\frac{4x^{12}+x^{6}y^{6}+x^{6}z^{6}}{6}+\sum_{sym}\frac{x^{12}+4x^{6}y^{6}+x^{6}z^{6}}{6}\ge \sum_{cyc}x^{10}yz+\sum_{sym}x^{7}y^{4}z$
which is just AM-GM
scorpius119
05.11.2006 23:11
We first show that
\[\frac{x^{9}+y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}\geq \frac{2}{3}x^{3/2}y^{3/2}\]
After clearing denominators, this becomes
\[3(x^{9}+y^{9})\geq 2 (x^{15/2}y^{3/2}+x^{9/2}y^{9/2}+x^{3/2}y^{15/2})\]
Since $x^{9}+y^{9}\geq 2x^{9/2}y^{9/2}$ by AM-GM and $x^{9}+y^{9}\geq x^{15/2}y^{3/2}+x^{3/2}y^{15/2}$ (because their difference is $(x^{15/2}-y^{15/2})(x^{3/2}-y^{3/2})\geq 0$), this inequality holds.
Applying similar inequalities to the other terms and adding them up, \[\frac{x^{9}+y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}+\frac{y^{9}+z^{9}}{y^{6}+y^{3}z^{3}+z^{6}}+\frac{z^{9}+x^{9}}{z^{6}+z^{3}x^{3}+x^{6}}\]
\[\geq \frac{2}{3}(x^{3/2}y^{3/2}+y^{3/2}z^{3/2}+z^{3/2}x^{3/2})\]
and the right hand side is greater than or equal to $2xyz=2$ by AM-GM.
paladin8
06.11.2006 00:48
Jan wrote: For $x,y,z > 0$ and $xyz=1$, prove that \[\frac{x^{9}+y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}+\frac{x^{9}+z^{9}}{x^{6}+x^{3}z^{3}+z^{6}}+\frac{y^{9}+z^{9}}{y^{6}+y^{3}z^{3}+z^{6}}\geq 2 \] What is the point of all the high powers? We can just substitute $a = x^{3}$, $b = y^{3}$, $c = z^{3}$ to get the same inequality with each power divided by $3$.
SplashD
06.11.2006 00:52
paladin8 wrote: Jan wrote: For $x,y,z > 0$ and $xyz=1$, prove that \[\frac{x^{9}+y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}+\frac{x^{9}+z^{9}}{x^{6}+x^{3}z^{3}+z^{6}}+\frac{y^{9}+z^{9}}{y^{6}+y^{3}z^{3}+z^{6}}\geq 2 \] What is the point of all the high powers? We can just substitute $a = x^{3}$, $b = y^{3}$, $c = z^{3}$ to get the same inequality with each power divided by $3$. That's what I did at first, but I like to work in integer powers and there would be fractional ones when homogenized. It's just one of my preferences.
Kurt Gödel
06.11.2006 01:36
Let $x^{3}= a$, $y^{3}= b$ and $z^{3}= c$. Then $abc = 1$ and \[\begin{aligned}\sum_{\textrm{cyc}}\frac{x^{9}+y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}& = \sum_{\textrm{cyc}}\frac{a^{3}+b^{3}}{a^{2}+ab+b^{2}}= \sum_{\textrm{cyc}}\left(\frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}}+2\frac{b^{3}}{a^{2}+ab+b^{2}}\right) \\ & = \sum_{\textrm{cyc}}(a-b)+2\sum_{\textrm{cyc}}\left(\frac{b^{3}}{a^{2}+ab+b^{2}}\right) = 0+2\sum_{\textrm{cyc}}\left(\frac{b^{4}}{b^{3}+a^{2}b+b^{2}a}\right)\\ & \geq 2\frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a^{3}+b^{3}+c^{3}+a^{2}b+b^{2}c+c^{2}a+ab^{2}+bc^{2}+ca^{2}}\\ & = 2\frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{\left(a^{2}+b^{2}+c^{2}\right)\left(a+b+c\right)}= 2\frac{a^{2}+b^{2}+c^{2}}{a+b+c}\end{aligned}\]So if $a^{2}+b^{2}+c^{2}\geq a+b+c$, we're done. But this is just Muirhead on $(2,0,0)\succ \left(\frac{4}{3},\frac{1}{3},\frac{1}{3}\right)$, because $abc = 1$.
Jan
06.11.2006 18:51
Nice solutions! I like the idea of using $(x^{3}-y^{3})+2y^{3}$. Here's my attempt. I hope it's correct In the first inequality, we use $a^{2}+b^{2}\geq 2ab$ In the second inequality, we use rearrangement: $a^{3}+b^{3}\geq \sqrt{ab}(a^{2}+b^{2})$ In the third inequality, we use am-gm. We prove that \[\sum_{cyc}{\frac{a^{3}+b^{3}}{a^{2}+ab+b^{2}}}\geq \sum_{cyc}\frac{a^{3}+b^{3}}{\frac{3}{2}(a^{2}+b^{2})}\geq \frac{2}{3}(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}) \geq 2(\sqrt[3]{abc})=2\]