Proof that $\frac{1+x^2}{n^2} \geq 1-e^{-x^2/n^2}$ for all $x,n \in \mathbb{R}$ I'm looking for a simple proof that $\frac{1+x^2}{n^2} \geq 1-e^{-x^2/n^2}$ for all $x,n \in \mathbb{R}$.
My first attempt was to express the exponential as a Taylor series:
$$\frac{1+x^2}{n^2} \geq \frac{x^2}{n^2}-\frac{1}{2!}\frac{x^4}{n^4}+\frac{1}{3!}\frac{x^6}{n^6}- \, ... \, .$$
Obviously 
$$\frac{1+x^2}{n^2} \geq \frac{x^2}{n^2},$$
so if I can show 
$$-\frac{1}{2!}\frac{x^4}{n^4}+\frac{1}{3!}\frac{x^6}{n^6}- \, ... <0,$$
then I'm done. But I'm stuck here, and also wondering if there's an even simpler way.

Set $y=x^2/n^2$. Then you want to show that
$$
\frac{1}{n^2}+y\ge 1-e^{-y}
$$
Note that $y\ge0$. A standard process is to consider
$$
f(y)=\frac{1}{n^2}+y-1+e^{-y}
$$
and note that $f(0)=1/n^2>0$. Also
$$
f'(y)=1-e^{-y}=\frac{e^y-1}{e^y}>0
$$
for $y>0$. Therefore the function $f$ is strictly increasing over $[0,\infty)$ and so
$$
f(y)>0
$$
for $y\ge0$.
