We prove this problem by induction. It is obvious when $n=1$. Suppose the statement is true for $n$, we will prove it is also true for $n+1$. In fact, by the Cauchy-Schwarz inequality, we have
$$\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n+1}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+\cdots+b_{n+1}^{2}\right) \geq\left(a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n+1} b_{n+1}\right)^{2}$$

By the assumption $\sum_{i=1}^{n+1} a_{i}^{2} \leq \sum_{i=1}^{n+1} b_{i}^{2}$, so $\sum_{i=1}^{n+1} b_{i}^{2} \geq \sum_{i=1}^{n+1} a_{i} b_{i}$. According to Abel's formula,
$$\begin{array}{l} 0 \leq \sum_{i=1}^{n+1} b_{i}\left(b_{i}-a_{i}\right)=\left(b_{1}-b_{2}\right)\left(b_{1}-a_{1}\right)+\left(b_{2}-b_{3}\right)\left(b_{1}+b_{2}-a_{1}-a_{2}\right)+\cdots \\ +\left(b_{n}-b_{n+1}\right)\left(\sum_{i=1}^{n} b_{i}-\sum_{i=1}^{n} a_{i}\right)+b_{n+1}\left(\sum_{i=1}^{n+1} b_{i}-\sum_{i=1}^{n+1} a_{i}\right) \end{array}$$

In the above sum, each term except the last one is non-positive (since $k \in\{1,2, \cdots, n\}$, we have $b_{k} \leq b_{k+1}$ and $\sum_{i=1}^{k} b_{i} \geq \sum_{i=1}^{k} a_{i}$, by the induction hypothesis). Therefore, we must have
$$b_{n+1}\left(\sum_{i=1}^{n+1} b_{i}-\sum_{i=1}^{n+1} a_{i}\right) \geq 0 \Leftrightarrow \sum_{i=1}^{n+1} b_{i} \geq \sum_{i=1}^{n+1} a_{i}$$

Note: The following result is due to Le Huu Dien Khue, which is a stronger result and can be directly obtained by Abel's formula (without induction)
Suppose $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1} \leq b_{2} \leq \cdots \leq b_{n}$ are real numbers, and satisfy
$a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2} \leq b_{1}^{2}+b_{2}^{2}+\cdots+b_{k}^{2} \quad k=1,2, \cdots, n$, prove
$$b_{1}+b_{2}+\cdots+b_{n} \geq \frac{a_{1}^{2}}{b_{1}}+\frac{a_{2}^{2}}{b_{2}}+\cdots+\frac{a_{n}^{2}}{b_{n}}$$