Proving inequality for $n\in \mathbb{N}$ How can we prove the following: For all $x>0 $ and $n\in \mathbb{N}$,
$$\frac{x^n}{1+x+x^2+...+x^{2n}}\leq\frac{1}{2n+1}.$$
I was wondering if someone can help me. Thanks.

As suggested by @fonfonx in the comments, you can write 
\begin{align}
\frac{x^n}{1+x+x^2+...+x^{2n}} &= \frac{1}{\frac{1}{x^n}+\frac{1}{x^{n-1}}+\ldots+\frac{1}{x}+1+x+\ldots+x^{n}}\\ 
 &= \frac{1}{1+(\frac{1}{x}+x)+(\frac{1}{x^2}+x^2)+\ldots+(\frac{1}{x^n}+x^n)}\\ 
 &\leq\frac{1}{1+2+2+\ldots+2}\\
 &=\frac{1}{1+2n}\\
\end{align}
where we have used the fact that $\frac{1}{x^i}+x^i \geq 2$ for $x>0$, which follows immediately from $(x^i-1)^2 \geq 0$.
