We are looking for a lower bound on the function, $\frac{1.31}{e^{\frac{1.31}{x+1}} - 1}$ for $x \geq 2$. This function seems to behave linearly. We believe that the following statement holds:
$$\frac{1.31}{e^{\frac{1.31}{x+1}} - 1} > x,$$ but we have not been able to prove it thus far.
Any hints or insights would be greatly appreciated. Any other lower bounds would also be useful.
Thanks!
On
$e^x \leq 1 + x + x^2$ for $x \leq 1$, so
$e^{\frac{1}{x+1}} \leq 1 + \frac{1}{x+1} + \frac{1}{(x+1)^2}$ for $x \geq 0$. Then
$e^{\frac{1}{x+1}} - 1 \leq \frac{1}{x+1} + \frac{1}{(x+1)^2}$
and $x \leq \frac{1}{\frac{1}{x+1} + \frac{1}{(x+1)^2}} \leq \frac{1}{e^{\frac{1}{x+1}} - 1}$
Bad stuff happens when you introduce the 1.31; for $\alpha >= 1$ you'll end up with
$\frac{x}{\alpha^2} \leq \frac{1}{\frac{\alpha}{x+1} + \frac{\alpha^2}{(x+1)^2}} \leq \frac{1}{e^{\frac{\alpha}{x+1}} - 1}$
with $\alpha = 1.31$. So this method doesn't produce an inequality as tight as you want.
As $x$ tends to infinity, the denominator of your expression (lets call it $f(x)$ ) goes to zero without ever being negative, so $f(x)\geq 0$ for all $x\geq 2$ and I hazard the guess that your function is monotone increasing.
The easiest way to show your assertion then is to show that the inequality holds at $x=2$, and then show that the derivative on the left side is always bigger than the derivative on the right side, which equals to $1$.
You can then bound $$f(2+t)\geq f(2)+t\cdot \inf_{t\geq 0} f'(2+t),$$ so if $\inf_{t\geq 0} f'(2+t)\geq 1$ this implies $$f(2+t)\geq f(2)+t \geq 2+t\quad \forall t\geq 0.$$