Solve the system of equation

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Randyyy

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Hey, the given task is as follows, solve the system of equation.

ekvsys.png

I tried solving it as I would a regular system of equation but it leads nowhere, I just got a nest of ln. I can see the obvious solution which is that x,y,z = 1 but I can´t figure out how to show it. This chapter is about derivatives so maybe that ties into it somehow?
 
The system is completely symmetric which leads one to try a solution where [MATH]x = y = z[/MATH]
Try that and see what you come up with.
 
Oh, okay, that makes so much sense, thank you! Just a quick question which might be a little stupid, but when can you conclude that they are symmetrical except that they look quite similar? Is there something in specific I can look for?
 
Randyyy said:
Oh, okay, that makes so much sense, thank you! Just a quick question which might be a little stupid, but when can you conclude that they are symmetrical except that they look quite similar? Is there something in specific I can look for?
Click to expand...
What are the numerical values of x, y & z?
 
Randyyy said:
Hey, the given task is as follows, solve the system of equation.

View attachment 23157

I tried solving it as I would a regular system of equation but it leads nowhere, I just got a nest of ln. I can see the obvious solution which is that x,y,z = 1 but I can´t figure out how to show it. This chapter is about derivatives so maybe that ties into it somehow?
Click to expand...
Without loss of generality, we can assume [MATH]0 < x \le y \le z.[/MATH]
[MATH]\text {ASSUME, for purposes of contradiction, that } x < y.[/MATH]
[MATH]\therefore ln(x) < ln(y) \implies 1 + ln(x) < 1 + ln(y) \implies z < y,\\ \text {which is false because } y \le z.[/MATH][MATH]\therefore x = y.[/MATH]
[MATH]\therefore x = 1 + ln(y) = 1 + ln(x) = z \implies x = y = z.[/MATH]
That is one way to prove Romsek’s observation.

[MATH]\therefore x = 1 + ln(z) \implies x = 1 + ln(x).[/MATH]
Now what?
 
JeffM said:
Without loss of generality, we can assume [MATH]0 < x \le y \le z.[/MATH]
[MATH]\text {ASSUME, for purposes of contradiction, that } x < y.[/MATH]
[MATH]\therefore ln(x) < ln(y) \implies 1 + ln(x) < 1 + ln(y) \implies z < y,\\ \text {which is false because } y \le z.[/MATH][MATH]\therefore x = y.[/MATH]
[MATH]\therefore x = 1 + ln(y) = 1 + ln(x) = z \implies x = y = z.[/MATH]
That is one way to prove Romsek’s observation.

[MATH]\therefore x = 1 + ln(z) \implies x = 1 + ln(x).[/MATH]
Now what?
Click to expand...

x = 1 +lnx can be rewritten as [MATH]x = e^{x-1} \implies x=1[/MATH]Just one question regarding this inequality [MATH]0 < x \le y \le z.[/MATH], how did you set it up? I get that all 3 of them has to be >0 because ln(x) is not defined for [MATH]x \leq 0[/MATH] but what about the rest? The proof itself I understand, thank you by the way!
 
Randyyy said:
x = 1 +lnx can be rewritten as [MATH]x = e^{x-1} \implies x=1[/MATH]Just one question regarding this inequality [MATH]0 < x \le y \le z.[/MATH], how did you set it up? I get that all 3 of them has to be >0 because ln(x) is not defined for [MATH]x \leq 0[/MATH] but what about the rest? The proof itself I understand, thank you by the way!
Click to expand...
I kind of cheated. There may be some high powered theorem that explicitly justifies what I did, but I cannot cite it.

But there are only six possibilities

[MATH]x \le y \le z,\\ x \le z < y,\\ y \ le x \le z,\\ y \le z < x,\\ z \le x \le y, \text { or} \\ z \le y < x.[/MATH]And it was obvious that the same argument would apply whichever possibility obtained. So pick one and show the argument. If someone wants to cavil, show them that it makes no difference which option you pick. I concede it is a handwaving approach.
 
Thanks a lot JeffM for taking your time to show me the proof! From what I could find this builds on the fact that something is symmetrical. But how could we know that it is symmetrical? What are key features that tells us "aha, there is symmetry here so I can go ahead and just do x=y=z" or like you did beautifully, set the proof up and show that it is indeed the case that x=y=z and that the only solution is 1?
 
I’d probably not use the word “symmetry.“ I’d see that the definitions are identical in form and that the definitions are circular: x is defined in terms of y, which is defined in terms of z, which is defined in terms of x.

That would make me suspect that there is an unstated but regular relationship among the unknowns. I have just seen enough of such problems to think about proving equality. I’m sorry not to be able to give a simple rule.
 
I think what you are saying makes perfect sense. Very insightful and educational. Was it enough to show that [MATH]x = e^{x-1}[/MATH] only has got 1 solution to conclude x,y,z only has got the solution of x,y,z = 1 or is it necessary to prove it in some other way?
 
I say this with some trepidation because I am not a professional mathematician and know nothing about proof theory.

Without some high powered theorem on what exactly defines symmetry and exactly what simplifications that permits, I suspect that a complete formal proof would need to start from something like

[MATH]\text {Let } a,\ b, \ c \in \mathbb R^+,\ 0 < a =min(x,\ y, \ z), \ b = med(x,\ y,\ z), \text { and } c = max(x,\ y, \ z).[/MATH]
Then I’d have to deal with 6 cases: a = x and b= y, a = x and b = z, a = y and b = x, a = y and b = z, a =. z and b = x, and a = z and b = y. You would then prove (just as I did for one case) that a = b = c for all six cases.

At that point, you would have to prove a = 1 is a valid solution. I’d probably do that via the Intermediate Value Theorem. Then I’d probably use implicit differentiation to show that the function is monotonically increasing for all positive a and therefore a = 1 is a unique solution. That probably is not the most elegant proof possible, but it works.
 
Thanks all for taking your time to assist on this task. Especially you JeffM with your thorough and extremely educational posts!
 
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