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I want to solve $e^{-x} < \frac{1}{x^2 + 1}$, assuming $x \in \mathbb{R}^+$. This can be solved via finding the solutions of $e^x>x^2+1$. This seems less troublesome, but I've found this inequality is very hard as well. I've tried everything I could think of, from logarithmic approaches to remove the exponential function to expressing $e^x$ as $\sum_{k=0}^\infty \frac{x^k}{k!}$. This last approach renders the inequality $-x^2 +\sum_{k=3}^\infty \frac{x^k}{k!} > 0$, which isn't any less tricky.

A last, desperate approach was to prove the last inequality via induction for $x \in \mathbb{N}$ and see if that could be extended to $\mathbb{R}$ somehow. Needless to say such long shot failed. How would one go about solving this inequality?

PD: This inequality is just a particular case of a more general inequality $e^{-\frac{|x|}{\alpha}}<\frac{\alpha}{x^2+\alpha}$, with $\alpha \in \mathbb{R}^+$. The problem presented above is this inequality under the assumptions $\alpha = 1, x \in \mathbb{R}^+$. Any thoughts on this more general, and far more complicated, problem are also welcomed -though are not my main interest.

Thanks in advance.

HallaSurvivor
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lafinur
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  • A definite integral approach: We have $x > \ln(x^2 + 1)$ for all $x > 0$ which follows from $$x - \ln(x^2 + 1) = \int_0^x \frac{(y - 1)^2}{1 + y^2},\mathrm{d} y.$$ – River Li Jun 29 '22 at 07:13

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Let us define $f(x) := e^{x} - x^{2} - 1$. Then $f(0) = 0$ and $f'(x) = e^{x} - 2x > 0$ for every real number (due to this result). In particular, $f(x) > 0$ for every $x > 0$ (because $f$ is strictly increasing on $\mathbb{R}_{>0}$).

Hopefully this helps!

Átila Correia
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