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So I was working on this question which states:

Let $x$, $y$, and $z$ be positive real numbers which satisfy the equation: $$ xy + yz + zx = xyz(x+y+z) $$

What is the maximum value for expressed as $\frac{A}{B}$? Return the sum $A+B$ $$ \frac{1}{(2x + y + z)^2} + \frac{1}{(x + 2y + z)^2} + \frac{1}{(x + y + 2z)^2} $$

My thought process was this:

$$xy + yz + zx = x^2yz + xy^2z + xyz^2$$

Then (I'm especially not sure about this step) let, $xy = x^2yz$, $yz=xy^2z$, and $zx=xyz^2$ and next take out a common factor:$y = xyz$, $z=xyz$ and $x=xyz$, this then implies that $x,y,z = 1$. Then plug in these two the top equation and simply, and you get: $\frac{3}{16}$ and $19$. I know that $19$ is the correct answer, but I don't know if it was because this bunny rabbit gives alternative reasoning that is much more complex:enter image description here

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  • $\begingroup$ the answer is correct.your is wrong! you don't know when is MAX. you can have a many choice as you want but it is not a proof. $\endgroup$ – chenbai Jun 2 '14 at 6:41
  • $\begingroup$ I'm sorry, but I don't understand your comment. $\endgroup$ – Dair Jun 2 '14 at 6:49
  • $\begingroup$ I mean the answer which you copied is correct. your idea is wrong. you can't set $xy=x^2yz$ etc and say you get $x=y=z=1$ which is the point of MAX.but if it is not your question, pls forget my comments. $\endgroup$ – chenbai Jun 2 '14 at 8:25
  • $\begingroup$ @chenbai: Yes, that is my question. Thank you, but do you have a particular reason why I can't do what I did? Like a counter example or something that illustrates the point? Thanks. $\endgroup$ – Dair Jun 2 '14 at 18:42
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let's have a example:

$x\ge 0,y \ge 0,z\ge0, x+y+z=1$, find max of $x^2y+y^2z+z^2x$

you may think $x=y=z=\dfrac{1}{3}$ is the point of max, but the real one is $x=\dfrac{2}{3},y=\dfrac{1}{3},z=0$ or its permutations.

you may think $x=y=z=1$ is min, but $x=1,y=z=0$ or its permutations are the min points.

so even $x=y=z$ are the solution in most cases, you have to prove it. sometime, it is a hard work, just like the sample that you post.

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  • $\begingroup$ Ok, this is what I was looking for! Thank you so much! $\endgroup$ – Dair Jun 3 '14 at 1:41

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