Solution:

Let $i$ be arbitrary and consider the set $A = \{a_{1}, a_{2}, \ldots, a_{n}\}$ defined by $a_{i} = s + 1 - x_{i}$ and let $a_{j} = 1 - x_{j}$ for all $j \neq i$. For example, if $i = 2$ then $A$ would be $\{1 - x_{1}, s + 1 - x_{2}, 1 - x_{3}, \ldots, 1 - x_{n}\}$. The AM-GM inequality on $A$ tells us

$$1 = \frac{(s + 1 - x_{i}) + \sum_{j \neq i}(1 - x_{j})}{n} \geq \left((1 + s - x_{i}) \prod_{j \neq i}(1 - x_{j})\right)^{\frac{1}{n}}$$

Which rearranges to give us

$$1 - (s + 1 - x_{i}) \prod_{j \neq i}(1 - x_{j}) \geq 0.$$

From here we can multiply both sides by $(1 - x_{i})$, then add $s$ to both sides and factorise the LHS to get:

$$(s + 1 - x_{i})\left(1 - \prod_{j = 1}^{n}(1 - x_{j})\right) \geq s.$$

Now multiply both sides by $\frac{x_{i}}{s(s + 1 - x_{i})}$ to get the following equation.

$$\left(1 - \prod_{j = 1}^{n}(1 - x_{j})\right) \frac{x_{i}}{s} \geq \frac{x_{i}}{s + 1 - x_{i}} \quad (1)$$

Note that this equation holds for all $i$. Now consider the sum of Equation 1 over all $1 \leq i \leq n$. Since $(1 - \prod (1 - x_{j}))$ is constant and $\sum \frac{x_{i}}{s} = 1$, the sum of all the LHS equals $\left(1 - \prod (1 - x_{j})\right)$. So we get

$$1 - \prod_{j = 1}^{n}(1 - x_{j}) \geq \sum_{i = 1}^{n} \frac{x_{i}}{s + 1 - x_{i}}$$

as required.