2 Find all real solutions of the system of equations
$$
\left\{\begin{array}{l}
5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right), \\
x y+y z+z x=1
\end{array}\right.
$$

(Supplied by Hua-Wei Zhu)
The equation (1) can be transformed into
$$
\frac{x}{5\left(1+x^{2}\right)}=\frac{y}{12\left(1+y^{2}\right)}=\frac{z}{13\left(1+z^{2}\right)}
$$
(3)
It is clear that $x, y, z$ have the same sign. First, we find the positive solutions.
There exist $\alpha, \beta, \gamma \in(0, \pi)$, such that $x=\tan \frac{\alpha}{2}, y=\tan \frac{\beta}{2}, z=\tan \frac{\gamma}{2}$,

Then
$$
\sin \alpha=\frac{2 x}{1+x^{2}}, \sin \beta=\frac{2 y}{1+y^{2}}, \sin \gamma=\frac{2 z}{1+z^{2}},
$$
(3) becomes
$$
\frac{\sin \alpha}{5}=\frac{\sin \beta}{12}=\frac{\sin \gamma}{13}.
$$
(2) can be transformed into
$$
\frac{1}{z}=\frac{x+y}{1-x y},
$$

which is
$$
\cot \frac{\gamma}{2}=\tan \frac{\alpha+\beta}{2}.
$$

Note that $z \neq 0, x y \neq 1$, because $\alpha, \beta, \gamma \in(0, \pi)$, so
$$
\frac{\alpha+\beta}{2}=\frac{\pi}{2}-\frac{\gamma}{2},
$$

which means $\alpha+\beta+\gamma=\pi$.

Thus, $\alpha, \beta, \gamma$ are the three interior angles of some triangle $ABC$. From (4) and the Law of Sines, we know that the ratio of the sides $a, b, c$ opposite to $\alpha, \beta, \gamma$ is $5:12:13$, so,
$$
\sin \alpha=\frac{5}{13}, \sin \beta=\frac{12}{13}, \sin \gamma=1.
$$

Thus, $x=\tan \frac{\alpha}{2}=\frac{1}{5}$ or $5, y=\tan \frac{\beta}{2}=\frac{2}{3}$ or $\frac{3}{2}, z=\tan \frac{\gamma}{2}=1$. Substituting $z=1$ into (2), it is easy to see that $x$ and $y$ are both less than 1. Therefore, $\left(\frac{1}{5}, \frac{2}{3}, 1\right)$ is the only positive solution.
Hence, the original system of equations has two solutions: $\left(\frac{1}{5}, \frac{2}{3}, 1\right)$ and $\left(-\frac{1}{5},-\frac{2}{3},-1\right)$.
Method 2 It is clear that $x, y, z$ have the same sign.
From (2), we get $x=\frac{1-y z}{y+z}$, substituting into (1) we get
$$
\begin{array}{l}
12\left(\frac{y^{2}+1}{y}\right)=5\left(\frac{1-y z}{y+z}+\frac{y+z}{1-y z}\right) \\
=5 \frac{(1-y z)^{2}+(y+z)^{2}}{(y+z)(1-y z)} \\
=5 \frac{\left(y^{2}+1\right)\left(z^{2}+1\right)}{(y+z)(1-y z)}, \\
\text { i.e., } 5\left(z^{2}+1\right) y=12(y+z)(1-y z), \\
\text { similarly, } 5\left(y^{2}+1\right) z=13(y+z)(1-y z). \\
\text { Simplifying, we get } \\
12 y^{2} z+17 y z^{2}=7 y+12 z, \\
18 y^{2} z+13 y z^{2}=13 y+8 z, \\
30 y z(y+z)=20(y+z), \\
5 y z(x+y)(x+z)=12 x z(y+x)(y+z) \\
=13 x y(z+x)(z+y). \\
\text { Let } \left\{\begin{array}{l}
x(y+z)=a, \\
y(z+x)=b, \\
z(x+y)
\end{array}\right. \\
\text { Then from (1), we get } a+b+c=2, \\
\text { (5) becomes } \quad 5 b c=12 c a=13 a b \\
\Leftrightarrow \frac{5}{a}=\frac{12}{b}=\frac{13}{c}=k. \\
\text { Substituting (7) into (6), } \frac{5}{k}+\frac{12}{k}+\frac{13}{k}=2, \\
\text { so } \quad k=15. \\
\text { Therefore, } \\
\left\{\begin{array}{l}
x y+x z=\frac{5}{15} \\
y z+x y=\frac{12}{15} \\
z x+z y=\frac{13}{15}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
y z=\frac{10}{15}, \\
z x=\frac{3}{15}, \\
x y=\frac{2}{15}.
\end{array}\right. \\
x y z= \pm \sqrt{\frac{10}{15} \cdot \frac{3}{15} \cdot \frac{2}{15}}= \pm \frac{2}{15}, \\
\text { Hence, the original system of equations has two solutions: }\left(\frac{1}{5}, \frac{2}{3}, 1\right) \text { and }\left(-\frac{1}{5},-\frac{2}{3},-1\right). \\
\text { Method 4 Let } 5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)=k, \text { then } \\
\frac{k}{x+\frac{1}{x}}=5, \frac{k}{y+\frac{1}{y}}=12, \frac{k}{z+\frac{1}{z}}=13. \\
\text { So } \frac{1}{\left(x+\frac{1}{x}\right)^{2}}+\frac{1}{\left(y+\frac{1}{y}\right)^{2}}=\frac{1}{\left(z+\frac{1}{z}\right)^{2}}, \\
\text { i.e., }\left(\frac{x}{x^{2}+1}\right)^{2}+\left(\frac{y}{y^{2}+1}\right)^{2}=\left(\frac{z}{z^{2}+1}\right)^{2}. \\
\text { Also, } x^{2}+1=x^{2}+x y+y z+z x=(x+y)(x+z),
\end{array}
$$

Adding the two equations, we get

So $y z=\frac{2}{3}, y=\frac{2}{3 z}$, substituting into (1) we get $z= \pm 1$.
Hence, the original system of equations has two solutions: $\left(\frac{1}{5}, \frac{2}{3}, 1\right)$ and $\left(-\frac{1}{5},-\frac{2}{3},-1\right)$.
Method 3
$x+\frac{1}{x}=\frac{x^{2}+x y+y z+z x}{x}=\frac{(x+y)(x+z)}{x}$,

So (1) can be transformed into
$$
\begin{array}{c}
\frac{x^{2}}{(x+y)^{2}(x+z)^{2}}+\frac{y^{2}}{(y+z)^{2}(y+x)^{2}}=\frac{z^{2}}{(x+z)^{2}(y+z)^{2}}, \\
\text { i.e., } \\
2 x^{2} y^{2}+2 x^{2} y z+2 x y^{2} z=2 x y z^{2}, \\
2 x y(x y+y z+z x)=2 x y z^{2}, \\
x y=x y z^{2}. \\
\text { Also, } x, y, z \neq 0, \text { so } z^{2}=1.
\end{array}
$$

So (8) can be transformed into