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.
Proving inequality for $n\in \mathbb{N}$
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real-analysis
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1This follows from the AM-GM inequality. – 2017-01-07
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0Hint: x+1/x>=2 if x>0 – 2017-01-07
2 Answers
1
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 $(x^i-1)^2 \geq 0$.
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From AM-GM we have $$\frac{1+x+\cdots + x^{2n}}{2n+1} \geq x^n $$ using the fact that $$1 + \cdots + 2n = \frac{(2n+1)2n}{2}.$$