Proof: Without loss of generality, assume $x_{1} \geq x_{2} \geq \cdots \geq x_{n}$. For $k \in\{1,2, \cdots, n\}$, let $b_{k}=\frac{1}{\sqrt{k}}$, we will first prove
$$2 \sum_{i=1}^{n} x_{i}^{2} \geq \sum_{i=1}^{n} x_{i} b_{i}$$

and
$$2 \sum_{i=1}^{n} x_{i} b_{i} \geq \sum_{i=1}^{n} b_{i}^{2}$$

By Abel's formula, we have
$$\begin{array}{l} \sum_{i=1}^{n} x_{i}\left(2 x_{i}-b_{i}\right)=\left(x_{1}-x_{2}\right)\left(2 x_{1}-b_{1}\right)+\left(x_{2}-x_{3}\right)\left(2 x_{1}+2 x_{2}-b_{1}-b_{2}\right)+\cdots \\ +\left(x_{n-1}-x_{n}\right)\left(2 \sum_{i=1}^{n-1} x_{i}-\sum_{i=1}^{n-1} b_{i}\right)+x_{n}\left(2 \sum_{i=1}^{n} x_{i}-\sum_{i=1}^{n} b_{i}\right) \end{array}$$

Since $x_{k} \geq x_{k+1} \quad \forall k \in\{1,2, \cdots, n\}$, we only need to prove $2 \sum_{i=1}^{n} x_{i} \geq \sum_{i=1}^{n} b_{i}$
By the given condition, we only need to prove
$$\sum_{i=1}^{k} \frac{1}{\sqrt{i}} \leq 2 \sqrt{k}$$

However, the last inequality is obviously true, because
$$\sum_{i=1}^{k} \frac{1}{\sqrt{i}} \leq \sum_{i=1}^{k} \frac{2}{\sqrt{i}+\sqrt{i-1}}=2 \sum_{i=1}^{k}(\sqrt{i}-\sqrt{i-1})=2 \sqrt{k}$$

This can also be derived by Abel's formula,
$$\begin{array}{l} \sum_{i=1}^{n} b_{i}\left(2 x_{i}-b_{i}\right)=\left(b_{1}-b_{2}\right)\left(2 x_{1}-b_{1}\right)+\cdots+\left(b_{n-1}-b_{n}\right)\left(2 \sum_{i=1}^{n-1} x_{i}-\sum_{i=1}^{n-1} b_{i}\right)+ \\ +b_{n}\left(2 \sum_{i=1}^{n} x_{i}-\sum_{i=1}^{n} x_{i}-\sum_{i=1}^{n} b_{i}\right) \end{array}$$
$b_{n} \geq b_{k+1}, \forall k \in\{1,2, \cdots, n\}$, so all terms are positive.