# System of equations (let me see You!)

1. Jan 5, 2007

### tehno

Consider the following system of algebraic equations:

$$2x+2y+2z-3=0$$
$$x^3+3yz^2-u=0$$
$$y^3+3x^2z-u=0$$
$$z^3+3xy^2-u=0$$

Problem questions:
a)Is the quadruple (1/2,1/2,1/2,1/2) the only solution to the system in $$\mathbb{R}_{+}$$ ?

b)If the answer to a) is "yes" how to proove that fact;if the answer is "no" how to determine how many other (nonelementar) solutions the system of equations like that will have ?

2. Jan 6, 2007

### Hurkyl

Staff Emeritus
This looks like homework. Have you tried anything?

3. Jan 7, 2007

### tehno

If you find this problem a homework and trivial than I'm really scared of level you work at.
No friend ,this is not a homework .
That's the challenge to you and everybody else.
(I know the answer just becouse I created that problem but whoever I asked failed to "solve" it so far.)

EDIT:I kindly suggest problem to be returned where It was posted first,or shifted to General math subforum.

Last edited: Jan 7, 2007
4. Jan 8, 2007

### dextercioby

I'll take a bite.

a) yes
b) cause Maple says so.

Daniel.

5. Jan 8, 2007

### tehno

Daniel your guess a) is correct but...
in relation to your remark b) may I ask you do you use computer to solve
say quadratic equations too?You know computers can do these tasks extremly well

I say there is a strict mathematical way to demonstrate the only positive solution to the system is the one I gave above.

Last edited: Jan 8, 2007
6. Jan 8, 2007

### Gib Z

Jesus techno you sick son of a *****, I expanded (x+y+z)^3 and all I've got it reduced to is $$u=\frac{9}{8} -(x^2y+2xyz+y^2z+xz^2)$$ and it took a bloody long time...anyone got a link to an expander?

7. Jan 9, 2007

### tehno

I don't want to spoil the party and I'll just give you two hints for this "homework" that might help:

1)The system is impossible to solve explicitely,by any kind of algebraic manipulation,substitution or transformation.
2)What is given is the only positive solution to it,and what should one notice is that there is certain level of symmetry of 3 variable function split among equations.

8. Jan 9, 2007

### J77

What a bizarre problem to create.

Still loks like homework to me... :tongue:

9. Jan 9, 2007

### tehno

Gee J77...
I would rather like to call me a sick b*****d and to try to do something like Gib Z,than to recognize dextor's most creative answer as the mathematically valid solution.

10. Jan 9, 2007

### AlephZero

I see the symmetry argument (and it's hard to write anything more about that without giving the complete answer, so I won't)

What I don't see is why you are restricting the solutions to positive quantities only. Why does that matter?

11. Jan 9, 2007

### Gib Z

haha techno, you love playing with peoples heads don't ya. You know when you said to dextor, the thing about computers solving quadratics, I thought that was a hint lol, i tried to reduce it to a quadratic for hours...thanks for telling me it was hopeless :D

12. Jan 9, 2007

### tim_lou

this is probably some kinda of olympiad-like problem. you probably need AM-GM, Cauchy inequalities or something like that.

from the first equation, you can easily show that 1/8>=xyz, somehow you need to show that (from the other equations):
xyz>=1/8 and the problem will be solved.

I'm not that fluid in these kinds of problem solving... I have barely done any problem in inequality besides reading the proofs for them... probably one of the reasons why I suck at Putnam...

perhaps one can approach this problem using a different method... consider, the Lagrange multiplier problem: extremizing the function (in positive reals):
$$f=xy^3+yz^3+zx^3$$
with the constraint:
$$C=x+y+z=3/2$$

and with the Lagrange multiplier being $u$

or you can consider assuming x>=y and get some contradiction.

Last edited: Jan 9, 2007
13. Jan 10, 2007

### tehno

tim_lou is on the right track.That's the best thinking so far!
One needs calculus and deeper insight to the problem.
That's what I expect of solvers here.

14. Jan 10, 2007

### Gib Z

O jesus Im not up to that kinda math yet lol.

15. Jan 11, 2007

### tehno

What kind of math you are not up to?Having learned the basics of multivariable calculus , by rewritting the equations a little,the proof of the claim can be presented in ~3 lines.
Another aspect of the problem is that it requires certain doseage of creativity.Unfortunately ,only one poster of this thread presented his creative side and gave pointers in the right direction...
FYI,I created this "homework " problem exploring the applications of:
http://cepa.newschool.edu/het/essays/theorem/euler.htm [Broken]

And I will not be giving any more hints than that.

Last edited by a moderator: May 2, 2017
16. Jan 11, 2007

### tim_lou

Using Lagrange multiplier... I bet one can minimize abc and show that the minimum is 1/2. but damn, that's like a 3 constraints, one function Lagrange multiplier problem... looks kinda messy.

by the way, I've just finished Calc three and have no idea what the heck Euler's Theorem is. looking from the link, it seems like it has something to do with vector space and eigenvalues and eigenvectors.... hmmm, linear Algebra.

I wonder if there is any elementary way of solving the system.

Last edited: Jan 11, 2007
17. Jan 11, 2007

### Hurkyl

Staff Emeritus
There is a purely algoritmic way to arrive at the answer. You can systematically eliminate variables until, say, x is the only one remaining: at which point you'll have a single polynomial equation in x which has only one positive root.

So, you substitute x = 1/2 and repeat the process.

In fact, I see how to do the elimination quickly:
Use the second equation to eliminate u from the others.
Use the first equation to eliminate z from the last two.
Compute a resultant of the remaining equations to eliminate y.
Voila! What remains is a polynomial in x.

Last edited: Jan 11, 2007
18. Jan 11, 2007

### tim_lou

that would probably require the use of cubic formula... with is ugly and inelegant.

edit: hmm maybe not, knowing that 1/2 is a solution, things should be reduced to quadratic...

19. Jan 11, 2007

### Hurkyl

Staff Emeritus
I wasn't even thinking that far ahead: you can eliminate variables simply by adding multiples of one equation to other equations.

I think I'm going to be lazy, though, and not bother with grinding through the algebra. I don't want to compute (3 - 2x - 2y)^3.

20. Jan 12, 2007

### Gib Z

I won't bother either, but since I already did (x+y+z)^3, Ill post that and someone can substitute in for us.

$$(x+y+z)^3=x^3+3xy^2+3x^2y+3x^2z+6xyz+y^3+3y^2z+z^3+3xz^2+3yz^2$$