Prove that $(1-x_{1})\cdots(1-x_{k}) \leq (1-\frac{1}{k})^{k}$ Let $x_{1},x_{2}, \ldots , x_{k}$ be positive real numbers such that
$$x_{1}+ x_{2}+ \cdots +x_{k}\geq1$$
$$0\leq x_{i}\leq1\text{ for }i \in \{1,2,\ldots ,k\}$$
I want to prove the following inequality
$$(1-x_{1})(1-x_{2})\cdots(1-x_{k}) \leq \left(1-\frac{1}{k}\right)^{k} \leq \frac{1}{e}$$
for any $k \geq 1$.
What is the best way to pursue in proving this?
 A: Let $x_i=1-y_i$ ($i\in\mathbb{N}\cap[1,k]$)
$$x_{1}+ x_{2}+ \cdots +x_{k}\geq1\implies \sum\limits_{i} y_i\leq k-1\implies AM=\frac{1}{k}\sum\limits_{i} y_i\leq \frac{k-1}{k}$$
$$GM=\sqrt[k]{\prod_i (1-x_i)}=\sqrt[k]{\prod_i y_i}$$
Using the AM-GM inequality, we obtain:
$$GM\leq AM\implies \prod_i (1-x_i)\leq \left(\frac{k-1}{k}\right)^k=\left(1-\frac{1}{k}\right)^k$$
A: Let $S=\sum_i x_k$, the AM-GM inequality gives
$$
\prod_i(1-x_i)\leq[(k-S)/k]^k\leq[(k-1)/k]^k=[1-1/k]^k.
$$
This gives the inequality in the title. For the second inequality, some hints are: use Bernoulli's inequality to prove that the sequence $\{(1-1/k)^k\}_k$ is increasing. The sequence is also bounded from above (say, by $1$) and, thus, converges to $\lim_{k\to\infty}(1+(-1)/k)^k=e^{-1}=1/e$.
A: You can do this using the AM-GM inequality: 

If $y_1,\dots,y_n\ge 0$, then 
  $$
\frac{y_1+\dots+y_n}{n}\ge \sqrt[n]{y_1\dots y_n}
$$

Applying this with $y_i=1-x_i$, we have:
\begin{align}
(1-x_1)\dots(1-x_k)&\le\left(\frac{(1-x_1)+\dots+(1-x_k)}{k}\right)^k\\
&=\left(\frac{k-(x_1+\dots+x_k)}{k}\right)^k\\
&\le\left(\frac{k-1}{k}\right)^k\\
&=\left(1-\frac1k\right)^k<\frac1e
\end{align}
