Prove that $\alpha_{n}=\frac{a_{1}+a_{2}+a_{3}+\cdots+a_{n}}{n}$ satisfies
$$\begin{aligned} \alpha_{n}^{2}-2 \alpha_{n} a_{n} & =a_{n}^{2}-2 \alpha_{n}\left[n \alpha_{n}-(n-1) \alpha_{n-1}\right] \\ & =(1-2 n) \alpha_{n}^{2}+2(n-1) \alpha_{n} \alpha_{n-1} \\ & \leqslant(1-2 n) \alpha_{n}^{2}+(n-1)\left(\alpha_{n}^{2}+\alpha_{n-1}^{2}\right) \\ & =-n \alpha_{n}^{2}+(n-1) \alpha_{n-1}^{2} \end{aligned}$$

Summing from $n=1$ to $N$, we have

that is $\square$
$$\sum_{n=1}^{N} \alpha_{n}^{2}-2 \sum_{n=1}^{N} \alpha_{n} a_{n} \leqslant-N \alpha_{N}^{2} \leqslant 0$$
$$\sum_{n=1}^{N} \alpha_{n}^{2} \leqslant 2 \sum_{n=1}^{N} \alpha_{n} a_{n}$$

Applying the Cauchy-Schwarz inequality to the right-hand side of the above equation, we get
$$\sum_{n=1}^{N} \alpha_{n}^{2} \leqslant 2 \sqrt{\left(\sum_{n=1}^{N} a_{n}^{2}\right)\left(\sum_{n=1}^{N} a_{n}^{2}\right)},$$

Dividing both sides of the above equation by $\sqrt{\sum_{n=1}^{N} a_{n}^{2}}$ and squaring, we get
$$\sum_{n=1}^{N} \alpha_{n}^{2} \leqslant 4 \sum_{n=1}^{N} a_{n}^{2}$$