3. In the domain of positive numbers, solve the system

$$
\begin{aligned}
& \sqrt{x y}+\sqrt{x z}-x=a, \\
& \sqrt{y z}+\sqrt{y x}-y=b, \\
& \sqrt{z x}+\sqrt{z y}-z=c
\end{aligned}
$$

where $a, b, c$ are given positive numbers.
SOLUTION. From the text of the problem, it follows that the unknowns $x, y, z$ are positive numbers, so the given system can be transformed into the following form

$$
\begin{aligned}
-\sqrt{x}+\sqrt{y}+\sqrt{z} & =\frac{a}{\sqrt{x}} \\
\sqrt{x}-\sqrt{y}+\sqrt{z} & =\frac{b}{\sqrt{y}} \\
\sqrt{x}+\sqrt{y}-\sqrt{z} & =\frac{c}{\sqrt{z}}
\end{aligned}
$$

If we add the equations of the previous system in pairs, we obtain the system

$$
\begin{aligned}
& \frac{b}{\sqrt{y}}+\frac{c}{\sqrt{z}}=2 \sqrt{x}, \\
& \frac{c}{\sqrt{z}}+\frac{a}{\sqrt{x}}=2 \sqrt{y}, \\
& \frac{a}{\sqrt{x}}+\frac{b}{\sqrt{y}}=2 \sqrt{z} .
\end{aligned}
$$

Further, after simple rearrangement,

$$
\begin{aligned}
& b \sqrt{z}+c \sqrt{y}=2 \sqrt{x y z}, \\
& c \sqrt{x}+a \sqrt{z}=2 \sqrt{x y z}, \\
& a \sqrt{y}+b \sqrt{x}=2 \sqrt{x y z} .
\end{aligned}
$$

By subtracting the first and third, and the second and third equations of the last system, we further obtain

$$
\begin{aligned}
b \sqrt{z}+(c-a) \sqrt{y} & =b \sqrt{x} \\
a \sqrt{z}-a \sqrt{y} & =(b-c) \sqrt{x} .
\end{aligned}
$$

We multiply both sides of the first equation of the previous system by the number $a$, and both sides of the second equation by the number $-b$. By adding both thus modified equations, we obtain

$$
a(b+c-a) \sqrt{y}=b(c+a-b) \sqrt{x},
$$

and similarly we also obtain

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

If for positive numbers $a, b, c$ the relation $b+c-a=0$ holds, then from the previous two equations it follows that $a+b-c=0, c+a-b=0$ as well. However, then $a=b=c=0$, which is not possible. Therefore, $b+c-a \neq 0$. From the last pair of equations, we express $\sqrt{y}$ and $\sqrt{z}$ in terms of $\sqrt{x}$ as follows:

$$
\begin{aligned}
& \sqrt{y}=\frac{b(c+a-b)}{a(b+c-a)} \sqrt{x} \\
& \sqrt{z}=\frac{c(a+b-c)}{a(b+c-a)} \sqrt{x}
\end{aligned}
$$

From this, we easily see that the expressions $b+c-a, c+a-b, a+b-c$ are all positive or all negative at the same time. By substituting $\sqrt{y}$ and $\sqrt{z}$ into the original system of equations, we obtain (after rearrangements) the solution $(x, y, z)$, where

$$
\begin{aligned}
& x=\frac{a^{2}(b+c-a)}{(c+a-b)(a+b-c)}, \\
& y=\frac{b^{2}(c+a-b)}{(b+c-a)(a+b-c)}, \\
& z=\frac{c^{2}(a+b-c)}{(c+a-b)(b+c-a)} .
\end{aligned}
$$

Given that we arrived at the solution of the system of equations exclusively through equivalent transformations, there is no need to perform a check.

The system has the above solution in the domain of positive numbers if and only if the following conditions are simultaneously satisfied: $b+c-a>0, c+a-b>0, a+b-c>0$, i.e., precisely when the positive numbers $a, b, c$ are the lengths of the sides of a triangle.

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