When does the inequality hold?

Question

I am trying to find a condition on $c$ such that the below inequality holds true

$$ \frac{1 - e^{-st}}{st} - \frac{1}{st+c} > 0 $$

where $s$, $c$ and $t$ are greater than $0$. I tried simpyfing it and got $c > (c + st) e^{-st}$, but I am not sure what do next.

Answer 1

We have

$$\frac{1 - e^{-st}}{st} - \frac{1}{st+c} > 0 \iff e^{-st}<1-\frac{st}{st+c}=\frac{c}{st+c}$$

$$\iff e^{st}>1+\frac{st}{c}$$

which is always true for $c\ge 1$.

To prove that, let consider $f(x)=e^x-1-\frac x c$ with $f(0)=0$ and show that

• for $c\ge 1 \implies \forall x>0\: f(x) >0 $

• for $0<c< 1 \implies \exists x_0>0 \:f(x_0) =0$

Answer 2

If you want to express c with s and t, you should maybe do the following:

$\frac{1-e^{-st}}{st}> \frac{1}{st+c}$ . Since $\frac{1}{st+c}>0$ and $\frac{1-e^{-st}}{st}>0$ we have:

$\frac{st}{1-e^{-st}}< st+c$

$\frac{st}{1-e^{-st}}-st<c$

I hope I answered correctly to your question.

Answer 3

Let $x = st >0$. Then your inequality rewrites:

$$\frac{1 - e^{-x}}{x} - \frac{1}{x+c} > 0 \quad \Leftrightarrow\quad e^x > \frac{1}{c}\ x + 1\; ;$$

since $x\mapsto e^x$ is strictly convex and $e^0=1$, you have $e^x \geq x + 1$ for all $x \in \mathbb{R}$ (with equality iff $x=0$) thus your inequality holds for all $x>0$ iff $\frac{1}{c} \leq 1$, i.e. iff $c \geq 1$.