Solve the system of equations

$$
\begin{gathered}
\frac{\sqrt{x+z}+\sqrt{x+y}}{\sqrt{y+z}}+\frac{\sqrt{y+z}+\sqrt{x+y}}{\sqrt{x+z}}=14-4 \sqrt{x+z}-4 \sqrt{y+z} \\
\sqrt{x+z}+\sqrt{x+y}+\sqrt{z+y}=4
\end{gathered}
$$
Solution. Let's introduce new unknowns: let $\sqrt{x+y}=a, \sqrt{y+z}=b, \sqrt{x+z}=c$. With this, our system of equations takes the following form:

$$
\begin{aligned}
\frac{c+a}{b}+\frac{b+a}{c} & =14-4 c-4 b \\
a+b+c & =4
\end{aligned}
$$

In equation (1), we can replace $c+a, b+a, -4c-4b$ with $4-b, 4-c, -4(4-a)$ respectively, according to equation (2). Therefore, after rearranging equation (1):

$$
\begin{aligned}
\frac{4-b}{b}+\frac{4-c}{c} & =14-4(4-a) \\
\frac{1}{b}+\frac{1}{c} & =a
\end{aligned}
$$

Substituting this into equation (2), we get:

$$
\left(b+\frac{1}{b}\right)+\left(c+\frac{1}{c}\right)=4
$$

The sum of a positive number and its reciprocal is at least 2, and it is exactly 2 if and only if the number is 1. Since $b$ and $c$ are positive, according to the obtained equation:

$$
\left(b+\frac{1}{b}\right)=\left(c+\frac{1}{c}\right)=2
$$

Thus, $b=c=1$, from which $a=4-b-c=2$. Squaring the original unknowns, we get the linear system of equations:

$$
x+y=4, \quad y+z=1, \quad x+z=1
$$

which has the unique solution:

$$
x=2, \quad y=2, \quad z=-1
$$

It is easy to verify that this triplet satisfies the equations of the problem.