Hallo :) I am hopeless with this exercise:

Solve the system of equations over the positive real numbers




where $a, b, c$ are positive real numbers.

I tried to +,- and / the equations with one another, but I din`t see any reasonable result. I also rised them to the power of two and then count,,, I had this solution


a firend of mine tried to count all three equations together and he got this

$2(\sqrt{xy}+...)-(x+...)=a +b+c$

$2(\sqrt{xy}+...)-((\sqrt x+\sqrt y+\sqrt z)^2-2(\sqrt {xy}+...))=a +b+c$

$4(\sqrt {xy}+...)-(\sqrt x+\sqrt y+\sqrt z)^2=a+b+c$

But we both don`t know what to do with that.

Do you know some reasonable method how to solve this system?

Thank you wery much!

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    $\begingroup$ Let $$u := \sqrt x \qquad \qquad \qquad v := \sqrt y \qquad \qquad \qquad w := \sqrt z$$ and write the system of equations as follows $$u v + u w - u^2 = a$$ $$u v + v w - v^2 = b$$ $$u w + v w - w^2 = c$$ Unfortunately, I don't know what to do with these. $\endgroup$ – Rodrigo de Azevedo Oct 8 '16 at 11:16

We can solve the system in the following way (though I'm not sure if it is "reasonable") :

We have $$\sqrt y+\sqrt z-\sqrt x=\frac{a}{\sqrt x}\tag1$$ $$\sqrt z+\sqrt x-\sqrt y=\frac{b}{\sqrt y}\tag2$$ $$\sqrt x+\sqrt y-\sqrt z=\frac{c}{\sqrt z}\tag3$$ From $(1)$, $$\sqrt z=\sqrt x-\sqrt y+\frac{a}{\sqrt x}\tag4$$

From $(2)(4)$, $$\sqrt x-\sqrt y+\frac{a}{\sqrt x}+\sqrt x-\sqrt y=\frac{b}{\sqrt y},$$ i.e. $$2\sqrt x-2\sqrt y+\frac{a}{\sqrt x}-\frac{b}{\sqrt y}=0$$ Multiplying the both sides by $\sqrt{xy}$ gives $$2x\sqrt y-2y\sqrt x+a\sqrt y-b\sqrt x=0,$$ i.e. $$y=\frac{2x\sqrt y+a\sqrt y-b\sqrt x}{2\sqrt x}\tag5$$

From $(3)(4)$, $$\sqrt x+\sqrt y-\left(\sqrt x-\sqrt y+\frac{a}{\sqrt x}\right)=\frac{c}{\sqrt x-\sqrt y+\frac{a}{\sqrt x}},$$ i.e. $$\frac{2\sqrt{xy}-a}{\sqrt x}=\frac{c\sqrt x}{x-\sqrt{xy}+a}$$ Multiplying the both sides by $\sqrt x\ (x-\sqrt{xy}+a)$ gives $$2x\sqrt{xy}-2xy+3a\sqrt{xy}-ax-a^2=cx,$$ i.e. $$y=\frac{2x\sqrt{xy}+3a\sqrt{xy}-ax-a^2-cx}{2x}\tag6$$

From $(5)(6)$, $$\frac{2x\sqrt y+a\sqrt y-b\sqrt x}{2\sqrt x}=\frac{2x\sqrt{xy}+3a\sqrt{xy}-ax-a^2-cx}{2x},$$ i.e. $$\sqrt y=\frac{a^2+(a-b+c)x}{2a\sqrt x}\tag7$$

From $(5)(7)$, $$\left(\frac{a^2+(a-b+c)x}{2a\sqrt x}\right)^2=\frac{(2x+a)\frac{a^2+(a-b+c)x}{2a\sqrt x}-b\sqrt x}{2\sqrt x},$$ i.e. $$\frac{(a^2+(a-b+c)x)^2}{4a^2x}=\frac{(2x+a)(a^2+(a-b+c)x)-2abx}{4ax}$$ Multiplying the both sides by $4a^2x$ gives $$(a^2+(a-b+c)x)^2=a((2x+a)(a^2+(a-b+c)x)-2abx),$$ i.e. $$x((a^2-b^2+2bc-c^2)x+a^3-a^2b-a^2c)=0$$ Finally, from $(7)(4)$, $$\color{red}{x=\frac{a^2(b+c-a)}{(a+b-c)(c+a-b)},\quad y=\frac{b^2(c+a-b)}{(b+c-a)(a+b-c)},\quad z=\frac{c^2(a+b-c)}{(b+c-a)(c+a-b)}}$$

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    $\begingroup$ @martina: (part1) I substituted $\sqrt z$, i.e. $(4)$ into $(2)$ and $(3)$. Then, I got two expressions $(5)(6)$ for $y$, so I was able to represent $\sqrt y$ by $x$ as written in $(7)$. Finally, substituting it into $(5)$ gives an equation on $x$. (part2) We can represent $x$ by $a,b,c$. Then, substituting it into $(7)$ and squaring the both sides give $y$ represented by $a,b,c$. Finally, substituting them into $(4)$ and squaring the both sides give $z$ represented by $a,b,c$ as I wrote. I hope this helps. $\endgroup$ – mathlove Oct 8 '16 at 17:43
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    $\begingroup$ @martina: I'm not sure if I answered all of your questions in the comments (I hope I did). If you have something unclear, let me know that. The reason why I did these steps is that I just noticed the method and in my opinion the calculations did not look very tedious. $\endgroup$ – mathlove Oct 8 '16 at 17:51
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    $\begingroup$ @martina: More concretely, finally, we get $x=\frac{a^2(b+c-a)}{(a+b-c)(c+a-b)}$. Substituting this into $(7)$ gives $\sqrt y$ represented by $a,b,c$. Then, squaring it gives $y$ represented by $a,b,c$. Now we have $x,y$ represented by $a,b,c$. Then, substituting these into $(4)$ gives $\sqrt z$, so finally square it to get $z$. The calculations are not so easy but the steps are simple. I hope this helps. $\endgroup$ – mathlove Oct 8 '16 at 18:16
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    $\begingroup$ @martina: We don't substitute $(4)$ into $(7)$. Firstly, we get $x=\frac{a^2(b+c-a)}{(a+b-c)(c+a-b)}$. Then, substituting this into $(7)$ gives $\sqrt y=\sqrt{\frac{b^2(c+a-b)}{(b+c-a)(a+b-c)}}$. Squaring the both sides gives $y=\frac{b^2(c+a-b)}{(b+c-a)(a+b-c)}$. Finally, substituting $x=\frac{a^2(b+c-a)}{(a+b-c)(c+a-b)}$ and $y=\frac{b^2(c+a-b)}{(b+c-a)(a+b-c)}$ into $(4)$ gives $\sqrt z=\sqrt{\frac{c^2(a+b-c)}{(b+c-a)(c+a-b)}}$. Squaring the both sides gives $z=\frac{c^2(a+b-c)}{(b+c-a)(c+a-b)}$. I hope this helps. $\endgroup$ – mathlove Oct 9 '16 at 4:36
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    $\begingroup$ @martina: $0$ is not a positive real number. So, after having $x((a^2-b^2+2bc-c^2)x+a^3-a^2b-a^2c)=0$, we can divide the both sides by $x\ (\not=0)$ to have $x=\frac{a^2(b+c-a)}{(a+b-c)(c+a-b)}$. $\endgroup$ – mathlove Oct 9 '16 at 4:45

Here are some ideas which, with some hindsight, save you the long computations.

Notice that your system of equations is cyclic in the variables (x,y,z) and (a,b,c). I.e. the next equation follows from the previous one by shifting all variables by one position ($x \to y$ and simultaneoulsy $a\to b$ etc.), where the last variable becomes the first. This calls for a solution which is also cyclic, i.e. once you have found $x=F(a,b,c)$, then $y=F(b,c,a)$ and $z = F(c,a,b)$.

Declare an unkown function $f$ and cylic $g,h$, i.e. $f=f(a,b,c)$, then $g=f(b,c,a)$ and $h = f(c,a,b)$.

Now you don't loose generality in writing your proposed solution as

$$ x = a^2 \frac{{f}}{{g} {h}}$$

which entails

$$ y = b^2 \frac{{g}}{{h} {f}} \quad ; \quad z = c^2 \frac{{h}}{{f} {g}}$$

Then your overall cyclic nature of $x,y,z$ is given.

Of course, here is where the hindsight came in. Since if you don't have a clue about the structure of the solution (see mathlove's answer), then you probably wouldn't choose such an ansatz. However, on plugging it into your equations things looks nice, since you get

$$ gh = b g + ch - af\\ hf = - b g + ch + af \\ fg = bg - ch + af $$

Again, this system is cyclic.

From the structure of these equations, $f,g,h$ can be chosen as a weigted sum of the constants $a,b,c$. Since the system is cyclic, we need to do this only for one equation. With unknown $A,B,C$, write $f = A a+ Bb + Cc$, then $g = A b+ Bc + Ca$, and $h = A c+ Ba + Cb$. The first equation is

$$ (A b+ Bc + Ca)(A c+ Ba + Cb) - b (A b+ Bc + Ca) - c(A c+ Ba + Cb) + a(A a+ Bb + Cc) = 0 $$

Sorting gives

$$ a^2(BC+A) + b^2(AC-A)+c^2(AB-A) + ab(AB+C^2-C+B) +bc(A^2+BC-B-C) + ac(AC+B^2-B+C) =0 $$

All the coefficients must vanish. $AB-A = 0$ gives $B=1$, then $AC-A = 0$ gives $C=1$, $BC+A = 0$ gives $A= -1$. Hence $A=-1$, $B=1$, $C=1$ and indeed all coefficients vanish.

This gives immediately the final solution.

The "calculation" part only consisted in the very easy determining of $A,B,C$.

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    $\begingroup$ thank you very much. your solution sounds interesting, but I don`t understand some steps. For example, how exactly do you know this $x=a^{2}\frac{f}{gh}$ ?? i hope, that when I understand this I would understand also how did you pugged those equations into my equations as well as all those equations bellow. $\endgroup$ – martina Oct 8 '16 at 17:45
  • $\begingroup$ @martina: Well, from my answer (further up), it is clear that x,y,z, should be cyclic functions in (a,b,c). So, in general, when designing a cyclic function, nobody forbids you to write is as $x = a^2 f/ (g h)$ with g and h being the cyclic versions of f. Of course, one would hardly do so expect in situations where one experiments a lot, has a brilliant guess or has hindsight (e.g. when knowing mathlove's solution already). Still, the rest of the discussion then shows that this setting allows for very easy computing of what f actually IS. [It's not checking that f = b+c-a holds.] $\endgroup$ – Andreas Oct 10 '16 at 12:59

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