Given $\lambda_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, n)$. Try to find all positive real solutions $x_{1}, x_{2}, \cdots, x_{n}$ of the equation
$$
\sum_{i=1}^{n} \frac{\left(x_{i}+\lambda_{i}\right)^{2}}{x_{i+1}}=4 \sum_{i=1}^{n} \lambda_{i}
$$
(Here $x_{n+1}=x_{1}$ ).
(Xiao Zhengang)
$$
\begin{array}{l} 
\sum_{i=1}^{n} \frac{\left(x_{i}+\lambda_{i}\right)^{2}}{x_{i+1}} \geqslant \frac{\left(\sum_{i=1}^{n} x_{i}+\sum_{i=1}^{n} \lambda_{i}\right)^{2}}{\sum_{i=1}^{n} x_{i}} \\
=\sum_{i=1}^{n} x_{i}+\frac{\left(\sum_{i=1}^{n} \lambda_{i}\right)^{2}}{\sum_{i=1}^{n} x_{i}}+2 \sum_{i=1}^{n} \lambda_{i} \\
\geqslant 2 \sqrt{\sum_{i=1}^{n} x_{i} \cdot \frac{\left(\sum_{i=1}^{n} \lambda_{i}\right)^{2}}{\sum_{i=1}^{n} x_{i}}}+2 \sum_{i=1}^{n} \lambda_{i} \\
=4 \sum_{i=1}^{n} \lambda_{i} .
\end{array}
$$

Equality holds if and only if $\frac{x_{1}+\lambda_{1}}{x_{2}}=\frac{x_{2}+\lambda_{2}}{x_{3}}=\cdots=\frac{x_{n-1}+\lambda_{n-1}}{x_{n}}=\frac{x_{n}+\lambda_{n}}{x_{n+1}}$, and $\sum_{i=1}^{n} x_{i}=\sum_{i=1}^{n} \lambda_{i}$.
By the property of proportional sequences, we get $x_{i}+\lambda_{i}=2 x_{i+1}$, i.e., $2 x_{i+1}-x_{i}=\lambda_{i}$, thus $2^{i} x_{i+1}-2^{i-1} x_{i}=2^{i-1} \lambda_{i}$, $i=1,2, \cdots, n$. Summing up, we get $\left(2^{n}-1\right) x_{1}=\sum_{i=1}^{n} 2^{i-1} \lambda_{i}$, so
$$
x_{1}=\frac{1}{2^{n}-1} \sum_{i=1}^{n} 2^{i-1} \lambda_{i} \text {. }
$$

By cyclic property, we get $x_{k}=\frac{1}{2^{n}-1} \sum_{i=1}^{n} 2^{i-1} \lambda_{k-1+i}, k=1,2, \cdots, n$. Where $\lambda_{n+k}=\lambda_{k}$.