40. First, take $x_{i}=\frac{\mathrm{F}}{2}(i=1,2, \cdots, n)$, substitute into $\sum_{i=1}^{n} x_{i} \geqslant C(n)\left(2 x_{i} x_{j}+\sqrt{x_{i} x_{j}}\right)$ to get $\frac{n}{2} \geqslant G(n) C_{n}^{2}\left(\frac{1}{2}+\frac{1}{2}\right)$

Then, $C(n) \leqslant \frac{\mathrm{F}}{n-1}$. Below, we prove that $C(n)=\frac{1}{n-1}$ satisfies the condition.
From $\left(1-x_{i}\right)+\left(1-x_{j}\right) \geqslant 2 \sqrt{\left(1-x_{i}\right)\left(1-x_{j}\right)} \geqslant 1(1 \leqslant i<j \leqslant n)$, we get $x_{i}+x_{j} \leqslant 1$

Taking the sum, we get $(n-F) \sum_{k=F}^{n} x_{k} \leqslant C_{n}^{2}$, i.e., $\sum_{k=1}^{n} x_{k} \leqslant \frac{n}{2}$. Therefore,
\begin{array}{l} \frac{1}{n-1} \sum_{1 \leqslant i<j \leqslant n}\left(2 x_{i} x_{j}+\sqrt{x_{i} x_{j}}\right)=\frac{1}{n-1}\left(2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}+\sum_{1 \leqslant i<j \leqslant n} \sqrt{x_{i} x_{j}}\right)= \\ \frac{1}{n-1}\left[\left(\sum_{k=1}^{n} x_{k}\right)^{2} - \sum_{k=1}^{n} x_{k}^{2}+\sum_{1 \leqslant i<j \leqslant n} \sqrt{x_{i} x_{j}}\right] \leqslant \\ 1-\left[\left(\sum_{k=1}^{n} x_{k}\right)^{2} - \sum_{k=1}^{n} x_{k}^{2} - \sum_{1 \leqslant i<j \leqslant n} 2 x_{i} x_{j} \leqslant\right. \\ \left.\left.\frac{1}{n-1}\left(\sum_{k=1}^{n} x_{k}\right)^{2} - \sum_{k=1}^{n} x_{k}^{2} + \sum_{1 \leqslant i<j \leqslant n} \sqrt{x_{i} x_{j}}\right]= \end{array}
$$\begin{array}{l} \left.\frac{1}{n-1} \frac{n-1}{n}\left(\sum_{k=1}^{n} x_{k}\right)^{2}+\frac{n-1}{2} \sum_{k=1}^{n} x_{k}\right] \\ \frac{1}{n}\left(\sum_{k=1}^{n} x_{k}\right)^{2}+\frac{1}{2} \sum_{k=1}^{n} x_{k} \leqslant \frac{1}{n}\left(\sum_{k=1}^{n} x_{k}\right) \cdot \frac{n}{2}+\frac{1}{2} \sum_{k=1}^{n} x_{k} \\ \sum_{k=1}^{n} x_{k} \end{array}$$

Thus, the original inequality holds.
Therefore, the maximum value of $C(n)$ is $\frac{1}{n-1}$.