<s>[INST] System of equations involving square roots Hallo :) I am hopeless with this exercise:

Solve the system of equations over the positive real numbers
$$\sqrt{xy}+\sqrt{xz}-x=a$$
$$\sqrt{zy}+\sqrt{xy}-y=b$$
$$\sqrt{xz}+\sqrt{yz}-z=c$$
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
$z+\sqrt{yz}-\sqrt{xy}-x=\frac{a^{2}}{2x}-\frac{c^{2}}{2z}$
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!
 [/INST] 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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