# Proving this algebra question!

1. Jun 9, 2013

### sam300

Hey guys, I am quite new to this forum, and just encountered this seemingly simple problem but still was not successful in solving it.

If,

$$\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$$

then prove that,

$$x=y=z$$

where x,y and z are real numbers.

I tried everything I could from basic identities to substituting x=a+y and x=b+z which one of my friend suggested but no luck yet.
The 'prove that' part seems obvious but I cannot make out how to write a proof.

I have been thinking and trying different ways to solve this for past 3 hours.
I would appreciate if you guys can help me out with this.

Last edited: Jun 9, 2013
2. Jun 9, 2013

### Integral

Staff Emeritus
Try rearranging to look like this:

$$\frac{x^2}{y^2}-\frac{x}{y}+\frac{y^2}{z^2}-\frac{y}{z}+\frac{z^2}{x^2}-\frac{z}{x}=0$$

Now think about how this could be true.

This is a homework style question, should have been posted in the homework forum. I am moving it there. You need to show us some effort.

3. Jun 9, 2013

### sam300

Hmm, if we pair the terms like this,
$$(\frac{x^2}{y^2}-\frac{x}{y})+(\frac{y^2}{z^2}-\frac{y}{z})+(\frac{z^2}{x^2}-\frac{z}{x})=0$$
We can see that $$\frac{x^2}{y^2}≥\frac{x}{y}$$
even if x or y are negative
that implies,
$$\frac{x^2}{y^2}-\frac{x}{y}$$
will either be positive or zero.
But if either of the pairs is positive then sum cannot be zero.
Therefore, $$\frac{x^2}{y^2}-\frac{x}{y}=0$$
which further implies $$\frac{x^2}{y^2}=\frac{x}{y}$$
which will give $$x=y$$ and eventually $$x=y=z$$

Is this correct or I am going wrong somewhere?

4. Jun 9, 2013

### Office_Shredder

Staff Emeritus

x=1, y=2?

5. Jun 9, 2013

### sam300

Oh, yea, then how else do I solve that?

Last edited: Jun 10, 2013
6. Jun 9, 2013

### Ray Vickson

You have an equation of the form $t_1 + t_2 + t_3 = 0,$ where
$$t_1 = \frac{x^2}{y^2} - \frac{x}{y}, \: t_2 = \frac{y^2}{z^2} - \frac{y}{z}, \: t_3 = \frac{z^2}{x^2} - \frac{z}{x}$$
There are several cases:
(1) All three t_i equal zero.
(2) One of the t_i = 0, one is > 0 and one is < 0.
(3) Two of the t_i are > 0 and one is < 0 (or two are < 0 and one is > 0).
Look at what happens in each case.

7. Jun 9, 2013

### CAF123

Is it a correct observation to note that $t_1, t_2, t_3$ are roots of a cubic polynomial with a zero x^2 coefficient? (not that I can see how(or even if) this helps be it the case).

8. Jun 9, 2013

### sam300

I don't think that will get me anything, case (2) and (3) can be zero sometimes or sometimes it may not.
I don't see how we can prove that in case (2) and (3) $t_1 + t_2 + t_3 ≠ 0,$

9. Jun 10, 2013

### Ray Vickson

What are you talking about? You are told that t_1 + t_2 + t_3 = 0. So, can you have that and cases (2) or (3) as well (remembering also the very special forms of the t_i)? You need to sit down and work it out carefully; don't just "guess".

10. Jun 10, 2013

### sam300

well, I thought I needed to prove t_1 = t_2= t_3 = 0 to get x/y=x^2/y^2 which will get me x=y
and to get that I need to rule out case (2) and (3) as a possibility.
But you seem to be saying something else. Okay, let me think.

11. Jun 10, 2013

### Ray Vickson

No, no, no. I am not saying that cases (2) or (3) can occur; however, we need to have some good reasons to eliminate them. One way would be to get a contradiction: assume case (2) and find out eventually that this leads to something going wrong. Ditto for case (3).

12. Jun 10, 2013

### sam300

Well I tried that and got inequalities like
$$\frac{x}{y}>=\frac{x^2}{y^2}$$

but simplifying them isn't getting me anywhere. Or maybe I did not get the thing required to find case (2) and (3) invalid.
Sorry, I'm not good at simplifying inequalities of variables.

13. Jun 11, 2013

### haruspex

Just checking... the RHs is definitely correct, yes?
$$\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$$
It isn't, perhaps,
$$\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}=\frac{y}{x}+\frac{z}{y}+\frac{x}{z}$$
?
I ask because that seems much easier!

14. Jun 11, 2013

### sam300

RHS is correct? ofc it is correct, the first equation is given to be true.
And I don't see how I could show eqn (2) to be incorrect.
And even if it is incorrect, how should help in proving?
Can you explain a bit in detail? I did not quite get you.

15. Jun 11, 2013

### Ray Vickson

He is asking you whether you might have copied down the original problem incorrectly, because the one you wrote is harder than the one he wrote.

16. Jun 11, 2013

### sam300

Oh, okay, no I 'm sure I wrote down the problem correctly.

I think I found a another method, but it requires a,b,c>0, can that be proved somehow?