Example 4 Find
$$
\left\{\begin{array}{l}
2 x^{3}=2 y\left(x^{2}+1\right)-\left(z^{2}+1\right), \\
2 y^{4}=3 z\left(y^{2}+1\right)-2\left(x^{2}+1\right), \\
2 z^{5}=4 x\left(z^{2}+1\right)-3\left(y^{2}+1\right)
\end{array}\right.
$$

the positive real solution $(x, y, z)$.
Solve: Since $x^{2}-2 x+1 \geqslant 0$, we have
$$
\begin{array}{l}
2 x^{3} \geqslant 2 x^{3}+2 x-x^{2}-1 \\
=(2 x-1)\left(x^{2}+1\right),
\end{array}
$$

When $x=1$, the equality holds;
$$
\begin{array}{l}
2 y^{4}+y=y^{4}+y^{4}+y \geqslant 3 y^{3} \\
\Rightarrow 2 y^{4} \geqslant 3 y^{3}-y-2\left(y^{2}-2 y+1\right) \\
=(3 y-2)\left(y^{2}+1\right),
\end{array}
$$

When $y=1$, the equality holds;
$$
\begin{array}{l}
2 z^{5}+2 z=z^{5}+z^{5}+z+z \geqslant 4 z^{3} \\
\Rightarrow 2 z^{5} \geqslant 4 z^{3}-2 z-3\left(z^{2}-2 z+1\right) \\
=(4 z-3)\left(z^{2}+1\right),
\end{array}
$$

When $z=1$, the equality holds.
$$
\begin{array}{l}
\text { Hence }\left\{\begin{array}{l}
2 y\left(x^{2}+1\right)-\left(z^{2}+1\right) \geqslant(2 x-1)\left(x^{2}+1\right), \\
3 z\left(y^{2}+1\right)-2\left(x^{2}+1\right) \geqslant(3 y-2)\left(y^{2}+1\right), \\
4 x\left(z^{2}+1\right)-3\left(y^{2}+1\right) \geqslant(4 z-3)\left(z^{2}+1\right)
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
2(y-x)\left(x^{2}+1\right)+(x-z)(x+z) \geqslant 0, \\
3(z-y)\left(y^{2}+1\right)+2(y-x)(y+x) \geqslant 0, \\
4(x-z)\left(z^{2}+1\right)+3(z-y)(z+y) \geqslant 0 .
\end{array}\right.
\end{array}
$$

If $x=\max \{x, y, z\}$, by $x, y, z>0$, we know $z-x \leqslant 0$.
From equation (2), we know $z \geqslant y$.
From equation (1), we get
$$
\begin{aligned}
0 & \leqslant 2(y-x)\left(x^{2}+1\right)+(x-z)(x+z) \\
& \leqslant 2(z-x)\left(x^{2}+1\right)+(x-z)(x+z) \\
& =(z-x)\left(2 x^{2}+2-x-z\right) \\
& \leqslant(z-x)\left(2 x^{2}+2-2 x\right) \\
& =(z-x)\left(2\left(x-\frac{1}{2}\right)^{2}+\frac{3}{2}\right) \leqslant 0 .
\end{aligned}
$$

Then $2(y-x)\left(x^{2}+1\right)+(x-z)(x+z)=0$, which holds if and only if $x=y=z$.
Similarly, if
$$
y=\max \{x, y, z\} \text { or } z=\max \{x, y, z\}
$$

we can also get $x=y=z$.
In summary, $(x, y, z)=(1,1,1)$.