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let $x\ge y\ge z\ge 0$,show that $$\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge x^2+y^2+z^2$$

my try: $$\Longleftrightarrow x^3y^2+y^3z^2+z^3x^2\ge xyz(x^2+y^2+z^2)$$

math110
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  • In any case, you can eliminaty one variable by setting for exmaple $x=a\ y$ and $y=b\ z$, with $a,b\ge 1$. Or some variant of that. For these values, one ends up with needing to show $a^2+b+a^3b^2(b-1)-a(b^2+1)\ge0$. You can also try to find a factorization of that right hand term in your equation. I've seen a similar problem being solved by finding out that it's some weird product of factors such as $x+\tfrac{1}{x}$, which is shown to be bigger than $1$ or something. – Nikolaj-K Nov 07 '13 at 09:36
  • The choice of $x \ge y \ge z \ge 0$ actually matters, it's not true if you only assume $x \ge 0, y \ge 0, z \ge 0$. – DanielV Nov 07 '13 at 11:17
  • http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=561642& It's my problem. :) – Steven Sun Nov 09 '13 at 01:48

4 Answers4

5

let $x=z+u,y=z+v,\to u\ge v \ge0$

$x^3y^2+y^3z^2+z^3x^2- xyz(x^2+y^2+z^2)=(u^2-uv+v^2)z^3+3u^2vz^2+uv(u^2-v^2)z+3u^2v^2z+u^3v^2 \ge0$

chenbai
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4

let $$E(a,b,c)=a^3b^2+b^3c^2+c^3a^2-abc(a^2+b^2+c^2)$$ then $$2E(a,b,c)=\sum a^3(b-c)^2-\sum a^2(b^3-c^3)=\sum a^2(b-c)^2(a+c-b)\ge 0$$ because $$\sum a^2(b^3-c^3)=\sum a^2(b-c)^3$$

math110
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    The last relation is interesting, is it a corollary of a more general concept from the type of summation, or just one you found for this problem? – DanielV Nov 07 '13 at 14:19
2

The equation can be rewritten as:

$$x^2 \left(\frac y z - 1\right) + y^2 \left(\frac z x - 1\right) + z^2 \left(\frac x y - 1\right) > 0 $$

Then since I noticed that $x > 0, y > 0, z > 0$ is not sufficient to solve the problem, I did a change of variables to convert the problem into one of that form:

$$v = \frac x y - 1, v > 0$$ $$u = \frac y z - 1, u > 0$$

$$\frac {x^2} {y^2} \left(\frac y z - 1\right) + \left(\frac z x - 1\right) + \frac {z^2} {y^2} \left(\frac x y - 1\right) > 0 $$

$$(v+1)^2u + (v+1)^{-1}(u+1)^{-1} - 1 + (u + 1)^{-2}\left((v+1)(u+1)-1\right) > 0$$

Then I put it into a computer to simplify the fraction because no way I want to do that by hand...

$$\frac{\left( {u}^{3}+2\,{u}^{2}+u\right) \,{v}^{3}+\left( 3\,{u}^{3}+6\,{u}^{2}+4\,u+1\right) \,{v}^{2}+\left( 3\,{u}^{3}+5\,{u}^{2}+3\,u\right) \,v+{u}^{3}+{u}^{2}+u}{\left( {u}^{2}+2\,u+1\right) \,v+{u}^{2}+2\,u+1} > 0$$

That's my first time solving this kind of problem.

DanielV
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    Not that I want to discourage you to post this, but it doesn't really improve over my comment to the question. If you make the substitution and expand the result like you did, it doesn't matter that you start out expressing it in terms of $\tfrac{y}{z}-1$. I've tried several choices for subsitutions and I posted the one which results in the simplest inequality above. In fact I think one only needs to show that the worst case is obtained for $a=1$. – Nikolaj-K Nov 07 '13 at 12:11
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Another way. $$\sum_{cyc}\left(\frac{x^2y}{z}-x^2\right)=(y-z)^2\left(\frac{x}{z}+\frac{x}{y}-1\right)+(x-y)(x-z)\left(\frac{y}{z}+\frac{y}{x}-1\right)\geq0.$$

Michael Rozenberg
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