# Bounding $e^{-tx}$ for fixed $t > 0$

For $t > 0$, how would I figure out without making use of a calculator whether

$e^{-xt} \le \frac{1}{t^2x^n}$, $\forall x > 0$

For $n \in \mathbb{N}$?

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You want that $$e^{tx/n}\geq t^{2/n}\,x$$ holds for any $x>0$. My statement is that the best (greatest) positive constant $\alpha$ for which $$e^x\geq \alpha\,x$$ holds for any $x>0$ is $\alpha=e$. So we have $e^{tx/n}\geq\frac{et}{n}\,x$ for any positive $x$. Since the exponential function is a convex function, in order to have $$e^{tx/n}\geq t^{2/n}\,x \quad\forall x>0$$ it is sufficient (and needed) that $$t^{2/n} \leq \frac{et}{n},$$ or: $$e^n\,t^{n-2}\geq n^n.$$