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!

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 $(7)(4)$,
$$\color{red}{x=\frac{a^2(b+c-a)}{(a+b-c)(c+a-b)},\quad y=\frac{b^2(c+a-b)}{(b+c-a)(a+b-c)},\quad z=\frac{c^2(a+b-c)}{(b+c-a)(c+a-b)}}$$
