Proving the inequality $e^{-2x}\leq 1-x$ How do I prove the inequality $e^{-2x}\leq1-x$ for $0\leq x\leq1/2$?
 A: For example, $(1-x)(1+2x)=1+x-2x^2\geq 1$ so
$$ e^{2x}\geq 1+2x \geq \frac{1}{1-x} $$
A: By Taylor expansion, we have $$\ln \frac{1}{1-x} = \sum_{n=1}^{\infty} \frac{x^n}{n},$$ whose convergence radius is $R=1$. 
The equation above can also be achieved by integrating both sides of $$\frac{1}{1-x} = \sum_{n=0}^{\infty} {x^n} .$$
For $0\leq x\leq 1/2$ we have the following upper bound 
$$
\sum_{n=1}^{\infty }\frac{x^{n}}{n}\leq \sum_{n=1}^{\infty }\frac{1}{
n\left( 2^{n}\right) }=\ln 2 \leq 2.
$$
Therefore 
$$
-\ln \left( 1-x\right) =\sum_{n=1}^{\infty }\frac{x^{n}}{n}\leq
2x.
$$
The given inequality follows.
A: We may start out in an elementary manner with a very known result that comes from W. Sierpiński inequality or $(e^x)$ Taylor expansion, namely:
$$e^x\geq 1+x $$
We get that:
$$e^{-x}\leq\frac1{x+1};  \space e^{-2x}\leq\frac1{2x+1}\leq1-x,\space \space \space 0\leq{x}\leq\frac{1}{2}$$
The proof is complete. 
A: A slightly ugly but standard beginning calculus approach is to let 
$$f(x)=1-x-e^{-2x}.$$
We want to show that $f(x)\ge 0$ in the interval $[0,1/2]$. 
A first experimental step might be to use software to graph $y=f(x)$ as $x$ ranges over our interval.  If we have a high degree of trust in the graphing software, the picture tells us that $f(x)$ is quite likely to be $\ge 0$ in our interval.  
Certainty is better. Use the derivative to study the behaviour of $f(x)$. Note that $f(0)=0$, and
$$f'(x)=2e^{-2x} -1.$$
The derivative is positive at $x=0$, and is clearly decreasing. It reaches $0$ at $x=(\ln 2)/2\approx 0.3465$.  So $f(x)$ is increasing in the interval $[0,(\ln 2)/2]$, and decreasing from $(\ln 2)/2$ on. By $x=1/2$, $f(x)$ is about $0.13212$, and in particular still positive. 
Thus $f(x)$ is $\ge 0$ for $0 \le x\le (\ln 2)/2$, and $f(x) >0$ for $(\ln 2)/2 \le x\le 1/2$. It follows that $f(x)\ge 0$ on the whole interval $[0,1/2]$ (and somewhat beyond $1/2$). 
Comment: There are far better ways to prove the inequality. But let's stick to calculusy approaches. To cut down on the negatives, note that equivalently we want to show that $e^{2x}(1-x) \ge 1$ on our interval (we multiplied both sides by the positive number $e^{2x}$). Let $g(x)=e^{2x}(1-x)-1$. Then $g(0)=0$. Also, $g'(x)=(1-2x)e^{2x}$, so $g$ is increasing on the interval $[0,1/2]$, and we are finished. 
A: Let $f(x)=e^{-2x}-(1-x)$. We have:


*

*$f''(x)=4e^{-2x}$, which is always positive. Hence, $f$ is convex.

*$f(0)=0$ 

*$f\left(\dfrac{1}{2}\right)=\dfrac{1}{e}-\dfrac{1}{2}\leq0$.


Since $f$ is convex, its curve is always below $0$ on $\left[0,\dfrac{1}{2}\right]$. This proves that $e^{-2x}\leq 1-x$.
A: For $x > 0$, the Taylor series for $e^{-2x}$ is the alternating series $\sum_n {(-2x)^n \over n!} = 1 - 2x + 2x^2 -...$ If $x < {1 \over 2}$, the terms decrease in magnitude.  It's a fact about alternating series that in an alternating series with terms of decreasing magnitude, the overall sum is going to be less than what you get if you add finitely many terms, stopping with a positive term. So we have
$$e^{-2x} < 1 - 2x + 2x^2$$
But after a little simple algebra, for $x > 0$ the statement $1 - 2x + 2x^2 < 1 - x$ is equivalent to the statement that $x < {1 \over 2}$. So the inequality follows. (The endpoint case $x = {1 \over 2}$ can be gotten by just plugging in the value).  
