""Proof""check for an inequality involving the exponential function This is a trivial question but I am having trouble to see what wehther my argumentation is valid or if I did some sign error. I suspect I messed somethin up again.
I start with the inequality
$$
1- \exp(-f(t)^2) \leq Ct \leq \frac{1}{2}
$$
for $t \leq \frac{1}{2C}$, $C>0$ and $f$ continuous. What I want to get is some inequality like
$$
f(t)^2 \leq 2Ct.
$$
on some reasonable interval $(0,t_0)$.
To this end, I subtract 1 on both sides of the first equation, divide by -1 and take the logarithm on both sides. Then I get
$$
-f(t)^2 \geq \ln(1-Ct).
$$
again dividing by -1 gives
$$
f(t)^2 \leq -\ln(1-Ct).
$$
Now wolframalpha says that
$$
-\ln(1-Ct)<2 Ct
$$
whenever
$$
0<x<1/2 (W(-2/e^2) + 2).
$$
I don't know what the function $W$ is but it works for me. Can I now conclude that
$$
f(t)^2 \leq 2Ct
$$
on the interval $(0,min(\frac{1}{2C},1/2 (W(-2/e^2) + 2))$? Also if it is true, is there some inequality to justify wolframalphas findings? Is it just the Taylor expansion of $\ln$?
 A: You dropped $C$ somewhere in your calculations that you should not have. There is no way that in a function where $t$ only shows up multiplied by $C$, any upper limit is not going to involve $C$ as well. Yet your expression involving $W$ has no $C$ in it.

Let's look carefully at where the inequality $- \ln(1 - Ct) < 2Ct$ holds. First, we can rewrite it as $$0 < \ln(1 - Ct) + 2Ct$$
Let $g(t) = \ln(1 - Ct) + 2Ct$. Then $$g'(t) = 2C - \dfrac{C}{1-Ct}$$
Note that $g(0) = 0$ and $g'(0) = C > 0$. Further, setting $g'(0) = 0$ gives $t = \frac 1{2C}$, exactly the upper limit you had set. Thus $g'(t) > 0$ on the interval $\left(0, \frac 1{2C}\right)$, and so $g$ is strictly increasing on this interval. Since $g(0) = 0$, we have that $g(t) > 0$ on all of $\left(0, \frac 1{2C}\right)$. And since $g$ is still non-decreasing at $\frac 1{2C}$, we also have $$g\left(\frac 1{2C}\right) \ge g\left(\frac 1{4C}\right) > 0$$
Which means
$$0 < \ln(1−Ct)+2Ct\\ -\ln(1−Ct) < 2Ct\\f(t)^2 < 2Ct$$ on
$\left(0, \frac 1{2C}\right]$.
