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Let $n$ be a positive integer (i.e. $n \geq 1$) and $x \in \mathbf R_{\geq 0}$. I need to show that $ \frac{nx^{1/n}}{n e^x + \sin(nx)} \leq \frac{1}{e^x} $ holds.

I tried looking at the $n$-th power and using Bernoulli's inequality: $ \left( \frac{nx^{1/n}}{n e^x + \sin(nx)} \right)^n \left( \frac{e^{-x} x^{1/n}}{1 + e^{-x} n^{-1} \sin(nx)} \right)^n = \frac{e^{-nx} x}{(1 + e^{-x} n^{-1} \sin(nx))^n}. $ With Bernoulli we get $(1 + e^{-x} n^{-1} \sin(nx))^n \geq 1 + e^{-x} \sin(nx)$. So $ \frac{e^{-nx} x}{(1 + e^{-x} n^{-1} \sin(nx))^n} \leq \frac{e^{-nx} x}{1 + e^{-x} \sin(nx)} = \frac{e^{-n} x}{e^{x} + \sin(nx)}.$ But I don't see how I can estimate this last quantity further. Can anybody give me a hint?

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The stated inequality is false. That is, the inequality $\frac{nx^{\frac{1}{n}}}{ne^{x}+\sin\left(nx\right)}\leq\frac{1}{e^{x}}$ does not hold for integers $n$ and $x\in \mathbb{R}_{\geq 0}$. To see why, notice that the left hand side is

$\frac{x^{\frac{1}{n}}}{e^{x}+\frac{\sin\left(nx\right)}{n}}\geq\frac{x^{\frac{1}{n}}}{e^{x}+1}$ $\geq\frac{x^{\frac{1}{n}}}{2e^{x}}=\frac{x^{\frac{1}{n}}}{2}\cdot\frac{1}{e^{x}}.$ For your inequality to hold, we would need $\frac{x^\frac{1}{n}}{2}\leq 1$ for all $x\in \mathbb{R}_{\geq 0}$, which is absurd.

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    Thank you. I misunderstood the dominated convergence theorem. I thought I needed to estimate the sequence on its own limit, but apparently any integrable function will do... :)2012-12-18
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Take $x=\pi/2$ and $n=2$ for a counterexample.