Properties of weighted average

Question

Consider a set of real positive numbers $\{x_1,x_2,...,x_n\}$ all greater than a positive real number $s$. Consider $n$ positive weights $w_i$ s.t. $\sum_{i=1}^{n}w_i=1$. Show that $\frac{1}{n}\sum_{i=1}^{n}w_ix_i\geq s$.

Answer 1

$$\sum_{i=1}^{n}w_ix_i-s=\sum_{i=1}^{n}w_ix_i-s\sum_{i=1}^{n}w_i=\sum_{i=1}^{n}w_i(x_i-s)\ge 0.$$