Problem 1544

Prove: Let $c_{i}=\sqrt{i}-\sqrt{i-1}, a_{i}=\sqrt{x_{i}}$, then the inequality becomes
$$\left(a_{1} c_{1}+a_{2} c_{2}+\cdots+a_{n} c_{n}\right)^{2} \geq a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}$$

Assume $b_{1}, b_{2}, \cdots, b_{n}$ are positive numbers, and satisfy $b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=1$ and $\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{i=1}^{n} b_{i}^{2}\right)=\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2}$ (sequences $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ and $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ are proportional). We only need to prove
$$a_{1} c_{1}+a_{2} c_{2}+\cdots+a_{n} c_{n} \geq a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}$$

By Abel's formula, the above inequality can be transformed into
$$\begin{array}{l} \sum_{i=1}^{n} a_{i}\left(c_{i}-b_{i}\right) \geq 0 \Leftrightarrow\left(a_{1}-a_{2}\right)\left(c_{1}-b_{1}\right)+\left(a_{2}-a_{3}\right)\left(c_{1}+c_{2}-b_{1}-b_{2}\right)+\cdots \\ +\left(a_{n-1}-a_{n}\right)\left(\sum_{i=1}^{n-1} a_{i}-\sum_{i=1}^{n-1} b_{i}\right)+a_{n}\left(\sum_{i=1}^{n} a_{i}-\sum_{i=1}^{n} b_{i}\right) \geq 0 \end{array}$$

This is valid, because for $k=1,2, \cdots, n$, we have
$$\sum_{i=1}^{k} c_{i}-\sum_{i=1}^{k} b_{i}=\sqrt{k}-\sum_{i=1}^{k} b_{i} \geq \sqrt{k}-\sqrt{k\left(\sum_{i=1}^{k} b_{i}^{2}\right)} \geq 0$$