How to prove/show $1- (\frac{2}{3})^{\epsilon} \geq \frac{\epsilon}{4}$, given $0 \leq \epsilon \leq 1$? How to prove/show $1- (\frac{2}{3})^{\epsilon} \geq \frac{\epsilon}{4}$, given $0 \leq \epsilon \leq 1$?
I found the inequality while reading a TCS paper, where this inequality was taken as a fact while proving some theorems. I'm not a math major, and I'm not as sufficiently fluent in proving inequalities such as these (as I would like to be), hence I'd like to know, why this is true (it does hold for a range of values of $\epsilon$ from $0$ to $1$), and how to go about proving such inequalities in general.

On of the most helpful inequalities about the exponential is $e^t\ge 1+t$ for all $t\in\mathbb R$.
Using this,
$$ \left(\frac32\right)^\epsilon=e^{\epsilon\ln\frac32}\ge 1+\epsilon\ln\frac32$$
for all $\epsilon\in\mathbb R$. 
Under the additional assumption that $ -\frac1{\ln\frac32}\le \epsilon< 4$, multiply with $1-\frac\epsilon4$ to obtain
$$\begin{align}\left(\frac32\right)^\epsilon\left(1-\frac\epsilon4\right)&\ge  \left(1+\epsilon\ln\frac32\right)\left(1-\frac\epsilon4\right)\\&=1+\epsilon\left(\ln\frac32-\frac14\right)-\frac{\ln\frac32}{4}\epsilon^2\\&=1+\frac{\ln\frac32}{4}\epsilon\cdot\left(4-\frac1{\ln\frac32}-\epsilon\right).\end{align}$$
Hence $\left(\frac32\right)^\epsilon\left(1-\frac\epsilon4\right)\ge1$ and ultimately $1-\frac\epsilon4\ge \left(\frac23\right)^\epsilon$ for all $\epsilon$ with $0\le\epsilon\le 4-\frac1{\ln\frac32}\approx1.53$
