Show that product is larger than sum Let $x_i > 1$, $i=1,\ldots,n$ denote some number.
I need to show the following:
\begin{align}
\prod_{i=1}^n x_i > \sum_{i=1}^n x_i.
\end{align}
Is there an easy way to show it's true or is there a basic reference? 
Edit:
Let $x_i$, $i=1,\ldots,n$ denote a sequence of integers larger than unity, i.e., $x_i > 1$ for all $i \in \{1,\ldots,n\}$.
I need to show the following:
\begin{align}
\prod_{i=1}^n x_i \geq \sum_{i=1}^n x_i.
\end{align}
 A: If $x_1,x_2,\ldots,x_n$ are real numbers greater than $1$ (but as the proof below shows, the inequality below is true for all real numbers $x_1,x_2,\ldots,x_n\geq 2$), then $$\prod_{i=1}^n\,x_i\geq \sum_{i=1}^n\,x_i\tag{*}$$
need not hold for $n>1$.  Take $x_i:=1+t$ for some $t>0$ and for all $i=1,2,\ldots,n$, then $$\prod_{i=1}^n\,x_i-\sum_{i=1}^n\,x_i=(1+t)^n-n\,(1+t)=:f(t)\,.$$
Clearly, $\lim\limits_{t\to 0^+}\,f(t)=1-n< 0$.  Therefore, by choosing a very small $t>0$, we get a counterexample.
If $x_1,x_2,\ldots,x_n$ are integers greater than $1$ (or real numbers greater than or equal to $2$), then (*) indeed holds.  This can be easily proven by induction on $n$ and it follows from the simple inequality
$$ab\geq a+b$$
for all integers $a,b>1$ (or for real numbers $a,b\geq 2$).  The equality cases of (*) for $x_1,x_2,\ldots,x_n\in\mathbb{Z}_{>0}$ are 


*

*$n=1$ with arbitrary $x_1\in\mathbb{Z}_{>1}$, and

*$n=2$ with $x_1=x_2=2$.
If you do not like induction, note that (*) is equivalent to
$$\sum_{i=1}^n\,\frac{1}{\prod\limits_{j\in\{1,2,\ldots,n\}\setminus\{i\}}\,x_j}\leq 1\,.$$  As $x_1,x_2,\ldots,x_n\geq 2$, the previous inequality is true because
$$\sum_{i=1}^n\,\frac{1}{\prod\limits_{j\in\{1,2,\ldots,n\}\setminus\{i\}}\,x_j}\leq \frac{n}{2^{n-1}}\leq 1$$
for all $n=1,2,3,\ldots$.  (You also can prove $n\leq 2^{n-1}$ without induction, by showing that $2n< (1+1)^n=2^n$ using the binomial expansion for $n\geq 3$, or via a combinatorial argument.)
