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?