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.
 A: The function $f(x) = 1 - \left( \tfrac{2}{3} \right)^x$ is increasing and concave down on the interval $[0, 1]$.  One consequence of this fact is that the curve lies above the secant line connecting $(0, f(0)) = (0, 0)$ and $(1, f(1)) = (1, \tfrac{1}{3})$.  This is the line $y = \tfrac{1}{3} x$, so actually the slightly stronger inequality holds for all $0 \le \epsilon \le 1$:
$$
1 - \left( \frac{2}{3} \right)^\epsilon \ge \frac{\epsilon}{3}
$$
Here's a picture of the curve $y = f(x)$, the secant line $y = \tfrac{x}{3}$, and the line $y = \tfrac{x}{4}$.

A: Let 
$$f(x)=1-\left(\frac{2}{3}\right)^x$$
then 
$$f'(x)=-\left(\frac{2}{3}\right)^x\log\left(\frac{2}{3}\right)$$
so $f$ is a concave function on the interval $[0,1]$ since $f'$ is decreasing hence the curve of $f$ is below the tangent line at the point $(0,f(0))=(0,0)$ which has the slope $f'(0)=-\log\left(\frac{2}{3}\right)\approx0.404$ and above the secant line connecting $(0,f(0))=(0,0)$ and $(1,f(1))=(1,\frac{1}{3})$ so
$$0.33x\leq f(x)\leq 0.41 x\quad \forall x\in[0,1]$$
A: 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 $0\le\epsilon\le 4-\frac1{\ln\frac32}\approx1.53$
