Can we say that $\prod_{i=1}^n (1-x_i)\ge 1-\sum_{i=1}^n x_i,\ \forall n\in \mathbb{N},\ \forall x_i\in [0,1)$? 
Can we say that $$\prod_{i=1}^n (1-x_i)\ge 1-\sum_{i=1}^n x_i$$ for all $n\in \mathbb{N}$ and for all $x_i\in [0,1)$?

The statement is easy to see to be true for $n=2,3$. However, what to do for general $n\in \mathbb{N}$? I am having this feeling that this should be a very trivial/well studied thing, but I am afraid to say I do not know a name for this in this in literature. So, please refer me to appropriate literature if this is pretty well known.
I thought of a pretty trivial "proof" using induction, but I am not sure about the applicability of induction in this case. I am still providing the "proof" for completeness:
The statement is true for $n=1,2$. Let it be true for $k$. Then, $\forall,\ x_1,\cdots,\ x_k\in [0,1)$, $$\prod_{i=1}^k (1-x_i)\ge 1-\sum_{i=1}^k x_i$$ Then note that, for any $\{y_i\}_{i=1}^{k+1}\in [0,1)$, we have
$$
\prod_{i=1}^{k+1}(1-y_i)
    \ge \left(1-\sum_{i=1}^k y_i\right)(1-y_{k+1})
    \ge 1-\sum_{i=1}^{k+1}y_i
$$
hence the claim is established.
$\blacksquare$
Any help is appreciated.
 A: Probabilistic Proof
Let $E_1,E_2,\ldots,E_n$ be pairwise statistically independent events in the probability space $(\Omega,\mathcal{F},\mathbb{P})$ such that $\mathbb{P}\left(E_i\right)=x_i\in[0,1]$ for each $i=1,2,\ldots,n$.  Then, by subadditivity of $\mathbb{P}$, $$\mathbb{P}\left(\bigcup_{i=1}^n\,E_i\right)\leq \sum_{i=1}^n\,\mathbb{P}\left(E_i\right)=\sum_{i=1}^n\,x_i\,.$$
On the other hand, note that
$$\mathbb{P}\left(\bigcup_{i=1}^n\,E_i\right)=1-\mathbb{P}\left(\bigcap_{i=1}^n\,E_i^\complement\right)=1-\prod_{i=1}^n\,\left(1-x_i\right)\,.$$  Hence,
$$1-\prod_{i=1}^n\,\left(1-x_i\right)\leq \sum_{i=1}^n\,x_i\,,$$
which is equivalent to the required inequality.  The equality holds if and only if at most one of the $x_i$'s is nonzero. 
A: In fact a more general inequality holds 
$$\prod_{i=1}^n (1-x_i)^{a_i}\ge 1-\sum_{i=1}^n a_i x_i$$
for $x_i\in [0,1)$ and $a_i\geq 1$, $i=1,\dots,n$.
This is a Weierstrass type inequality (or Weierstrass product inequality).
See for example at page 71, Chapter III (Bernoulli's inequality) in Classical and New Inequalities by Mitrinovic, Pecaric, Fink.
P.S. One more reference: article
