Just apply Cauchy-Schwarz's ineq

By Jensen's applied to $\sqrt{x}$, we get $\sum_i \sqrt{x_i\left(\sum_{j\not=i} x_j\right)}\leq \sqrt{n\sum_{i\not=j} x_ix_j}$ But by AM-GM, we have that $n\sum_{i\not=j} x_ix_j\leq (n-1)\left(\sum_i x_i\right)^2$ so that $\sum_i \sqrt{x_i\left(\sum_{j\not=i} x_j\right)}\leq \sqrt{(n-1)\left(\sum_i x_i\right)^2}$ which is exactly the inequality we want to prove.

Applying C-S on the two sequences $\left\{\sqrt{x_{k}}\right\}$ and $\left\{ \sqrt{x_{n+1} -x_{k}}\right\}$ $$\sum_{k=1}^n x_{k} \sum_{k=1}^n (x_{n+1} -x_{k}) \ge \left(\sum_{k=1}^n\sqrt{ x_k (x_{n+1}-x_k)}\right)^2$$ The result follows$.\Box$