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Let $R>2$ be a real number. Then, for any $n\geq 1$, it holds that

$$ \frac{ \exp(n R)+ 1}{\exp( n R) - \exp(\pi R/2)} \leq \exp( R/n^2).$$

How do I prove this?

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For $n=1$ the inequality: $$e^R+1\leq e^R(e^R-e^{\pi R/2})$$ is false, because $e^{(1+\pi/2)R}\geq e^{2R}-e^R-1$ for $R\geq 0$. – Pacciu Dec 5 '11 at 16:19

The inequality as stated: $$ \frac{ \exp(n R)+ 1}{\exp( n R) - \exp(\pi R/2)} \leq \exp( R/n^2) $$ can not be true for all $R>2$ and $n \geq 1$. Indeed, choose $n=2$, then for all $2 < r < r_\ast$, the inequality does not hold, where $r_\ast$ is the unique positive root of $$ e^{2 r}-e^{9 r/4}+\exp\left({\frac{\pi r}{2}+\frac{r}{4}}\right)+1 $$ approximately equal $r_\ast \approx 2.118953$.

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