做个小题

Siyunshan

若正实数列 $\left\{x_{n}\right\}_{n \geq 1}$ 对任意正整数 $n$ 都有:
$$
\frac{1}{n}\left(x_{1}+x_{2}+\cdots+x_{n}\right) \geq\left(\frac{1}{n+1}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n+1}^{2}\right)\right)^{\frac{1}{2}}
$$
证明：这是一个常数列

Siyunshan

提示：考虑拉格朗日恒等式

xcx

\begin{aligned}
Q_{n+1}^{2}-A_{n+1}^{2} &=\frac{1}{(n+1)^{2}}\left[(n+1)\left(a_{1}^{2}+\cdots+a_{n+1}^{2}\right)-\left(a_{1}+\cdots+a_{n+1}\right)^{2}\right] \\
&=\frac{1}{(n+1)^{2}} \sum_{1 \leq i<j \leq n+1}\left(a_{i}-a_{j}\right)^{2} \\
& \geq \frac{1}{(n+1)^{2}}\left(\sum_{i=1}^{n+1}\left(a_{1}-a_{i}\right)^{2}+\sum_{i=1}^{n+1}\left(a_{m}-a_{i}\right)^{2}-\left(a_{1}-a_{m}\right)^{2}\right.\\
&=\frac{1}{(n+1)^{2}}\left[\sum_{i≠1, m}\left[\left(a_{1}-a_{i}\right)^{2}+\left(a_{m}-a_{i}\right)^{2}\right]+\left(a_{1}-a_{m}\right)^{2}\right] \\
& \geqslant \frac{1}{(n+1)^{2}}\left[\frac{n-1}{2}\left(a_{1}-a_{m}\right)^{2}+\left(a_{1}-a_{m}\right)^{2}\right] \\
&=\frac{1}{2(n+1)}\left(a_{1}-a_{m}^{2}\right), \forall 1 \leq m<n+1 \\
\end{aligned}
固定 $m$,我们有
\begin{aligned}
& \forall A_{n}^{2}=A_{n+1}^{2}+\frac{1}{2}\left(\frac{1}{n+1}+\frac{1}{n+2}\right)\left(a_{1}-a_{m}\right)^{2} \geqslant \cdots \geqslant \\
& A_{n+k}^{2}+\frac{1}{n+1}+\cdots+\frac{1}{n+k}{(a_{1}-a_{m})}^2
\end{aligned}
由调和级数发散知$a_{1}=a_{m}$