How to prove that $e^{1/x} < \frac{1}{x} + 1$ for $x \in \mathbb{R}_{+}$?

Question

I could reduce this problem to showing that there is no solution for $e^{1/x} = \frac{1}{x} + 1$ and both functions are continuous, but I am unable to prove the emptiness of the set of solutions either.

Answer 1

Standard by Maclaurin exp. that $e^y = 1+y \cdots > 1+y$. So put $y=1/x$ and reverse the inequality and then you'd be OK.

Answer 2

I guess you want to prove "$>$". Replace $1/x \mapsto y$, recall your first lecture on Analysis and you're done.

Answer 3

To show the reverse inequality just set

• $\ln t = 1/x$ for $t > 1$

So, you get $$t > \ln t + 1 \mbox{ for } t > 1$$ which follows immediately from $1-\frac{1}{t} > 0$ for $t>1$ and $1 = \ln 1 + 1$.