Prove that by embedding the factor $x_{2}+x_{3}+\cdots+x_{n}+x_{1}$ on the left side of the inequality, it is equivalent to embedding the factor $x_{1}+x_{2}+\cdots+x_{n}$. Applying the Cauchy-Schwarz inequality, we have:
$$\begin{array}{l} \left(\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}}+\frac{x_{n}^{2}}{x_{1}}\right)\left(x_{2}+x_{3}+\cdots+x_{n}+x_{1}\right) \geqslant \\ \left(\frac{x_{1}}{\sqrt{x_{2}}} \cdot \sqrt{x_{2}}+\frac{x_{2}}{\sqrt{x_{3}}} \cdot \sqrt{x_{3}}+\cdots+\frac{x_{n-1}}{\sqrt{x_{n}}} \cdot \sqrt{x_{n}}+\frac{x_{n}}{\sqrt{x_{1}}} \cdot \sqrt{x_{1}}\right)^{2}= \\ \left(x_{1}+x_{2}+\cdots+x_{n-1}+x_{n}\right)^{2} \end{array}$$

Thus,
$$\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}}+\frac{x_{n}^{2}}{x_{1}} \geqslant x_{1}+x_{2}+\cdots+x_{n-1}+x_{n}$$

Similarly, we can prove:
(1) Given that $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and their sum is 1, prove that: $\frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n-1}^{2}}{a_{n-1}+a_{n}}+\frac{a_{n}^{2}}{a_{n}+a_{1}} \geqslant \frac{1}{2}$. (24th All-Soviet Union Mathematical Olympiad Problem)
(2) $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n}$ are two sets of positive real numbers, and $\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} b_{k}$, prove that: $\sum_{k=1}^{n} \frac{a_{k}^{2}}{a_{k}+b_{k}} \geqslant \frac{1}{2} \sum_{k=1}^{n} a_{k}$. (1991 Asia-Pacific Mathematical Olympiad Problem)