Proving the Inequality: $ (x_{1}+x_{2}+...+x_{n})\leq n^{n} x_{1}x_{2} \cdots x_{n}$ I am stuck on proving the following inequality:
Let: $x_{1},x_{2},...,x_{n}\geq 0$. Prove that:$(x_{1}+x_{2}+...+x_{n})\leq n^{n} x_{1}x_{2}...x_{n}$ 
where $n$ is a natural number.
 A: For $n=2$, set $x_1=\frac 1 2$ and $x_2=\frac 1 4$ to see that, if your inequality held true, $$\frac 3 4 \le 4\cdot\frac 1 2 \cdot \frac 1 4=\frac 1 2$$ which is a contradiction.

So, you must have claimed that, $$(x_1+x_2 \cdots +x_n)^n \ge n^n (x_1 x_2 \cdots x_n)$$ for positive values of $x_i$ and for all $n \in \Bbb N$.
This nicely rearranges to give, 
$$\dfrac{x_1+x_2+\cdots+x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n}$$
Note that this is nothing but the assertion that $AM \ge GM$ where $AM$ stands for the arithmetic mean and $GM$ stands for the geometric mean. 
Note that for $n=2$, the proof of this inequality relies on the fact that square numbers are positive. Instead of my reproducing a proof here, I suggest you'd look into Wikipedia link.
Corollary: 
If the sum $x_1+x_2+x_3+ \cdots + x_n \le 1$, we have that $$x_1+x_2+ \cdots +x_n \ge (x_1+x_2+x_3+\cdots+x_n)^n \ge n^n x_1x_2\cdots x_n \\ \quad \\ \boxed{x_1+x_2+\cdots + x_n \ge n^n x_1x_2 \cdots x_n}$$ 
A: The inequality of the post, as currently stated, is not universally correct.
It fails immediately if some but not all the $x_i$ are $0$. But it also fails when all the $x_i$ are positive.  The failure is essentially automatic, by continuity, or because the inequality is not homogeneous.
Let $\displaystyle x_i=\frac{1}{n^2}$ for all $n$.
Then $\displaystyle\sum x_i=\frac{1}{n}$.
But $\displaystyle\prod x_i=\frac{1}{n^{2n}}$, so $\displaystyle n^n\prod x_i=\frac{1}{n^n}$.
The right-hand side, when $n>1$, is less than the left-hand side.
If the inequality is to hold, conditions other than positivity must be put on the $x_i$.
A: False as said by others, an easy counterexample is reached with $x_1=77$ and $x_k=0$ for $k=2,\ldots n$. 
