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.
 A: For any $x \geq 0$ we have $1-e^{-x} \leq x$ $\,\,$ (1). Hence $1-e^{-x^{2}/n^{2}} \leq x^{2}/n^{2}$ which gives the inequality you want. To prove (1) consider $1-e^{-x} -x$. Its derivative is $e^{-x}-1$ which is negative. Since the function vanishes at $0$ and is decreasing it must be $\leq 0$ on $[0,\infty)$.
A: It is $e^t \geq t + 1$ for all $t \in \mathbb R$ (easily proven by elementary calculus), thus :
$$e^{-x^2/n^2} \geq 1 - x^2/n^2$$
But $1/n^2 >0$ for all $n \in \mathbb R$, thus if you add it to the LHS it will still be greater or equal to the RHS :
$$e^{-x^2/n^2} + 1/n^2  \geq 1 - x^2/n^2 \Leftrightarrow x^2/n^2 + 1/n^2 \geq 1-e^{-x^2/n^2} $$
$$\Leftrightarrow$$
$$\boxed{\frac{1+x^2}{n^2} \geq 1 - e^{-\frac{x^2}{n^2}}}$$
A: 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 $y\ge0$.
