Show that $e^{-x}/(1-x) \leq e^{x^2}$ for some interval around $0$ Show that  $e^{-x}/(1-x) \leq e^{x^2}$ for some interval around $0$.
What I've tried
My first attempt was to expand the LHS by applying a taylor expansion to both terms and then trying to bound the terms to get rid of one of the summations. I think this might work, but I am missing a trick on how to proceed.
My other thought is to choose two points on either side of $0$ such that the difference of the LHS and RHS is nonpositive at these points (by trial and error). Then I can show that if the difference is ever positive, then there must be some maximum in the interval. Taking the derivative to find all critical points in this interval and showing that they lead to a nonpositive result shows the claim. Unfortunately, the equation I get by setting the derivative to 0 involves some terms I do not know how to handle without the Lambert function.
I am looking for hints on how to proceed.
 A: This rearranges to
$$\frac1{1-x}\leq e^{x+x^2}.$$
The right side is, using Taylor series,
$$1+(x+x^2)+\frac{(x+x^2)^2}2+O(x^3)=1+x+\frac32x^2+O(x^3),$$
while the left side is $1+x+x^2+O(x^3)$. Subtracting, the inequality becomes $\frac12x^2\geq O(x^3)$, which holds for small enough $x$.
A: Consider the function
$$f(x)=\frac{e^{x^2}}{\frac{e^{-x}}{1-x}}=e^{x^2+x}(1-x)$$
We can compute $f'(x)=e^{x^2+x}(x-2x^2)$ and $f''(x)=e^{x^2+x}(-4x^3-x+1)$ to see that $f'(0)=0$ and $f''(0)=1>0$. Thus, $f$ has a local minimum at $0$, so for some open interval $I$ containing $0$,
$$f(x)\geq f(0)=e^{0^2+0}(1-0)=1\text{ for all }x\in I$$
It follows that $\frac{e^{x^2}}{\frac{e^{-x}}{1-x}}\geq 1$ on $I$, that is,
$e^{x^2}\geq\frac{e^{-x}}{1-x}$.
A: Consider two cases based on the sign of $x$:
Suppose $x$ is positive:
The power series expansion of $e^x$ is $\sum_{i=0}^\infty \frac{x^i}{i!}$.  Therefore, the power series expansion of $e^{x^2}$ is $\sum_{i=0}^\infty\frac{x^{2i}}{i!}$.  Note that for any $n$,
$$
\sum_{i=0}^n \frac{x^{2i}}{i!}\leq e^{x^2}.
$$
The power series for $e^{-x}$ is $\sum_{i=0}^\infty \frac{(-x)^i}{i!}$, and it is alternating (for $x$ sufficiently close to $0$).  Therefore, whenever $m$ is even,
$$
\sum_{i=0}^{m}\frac{(-x)^i}{i!}\geq e^{-x}.
$$
Therefore, it is enough to find an $n$ and even $m$ so that
$$
\sum_{i=0}^{m}\frac{(-x)^i}{i!}\leq (1-x)\sum_{i=0}^n \frac{x^{2i}}{i!}.
$$
When $n=1$ and $m=2$, we have the inequality
$$
1-x+x^2/2\leq(1-x)(1+x^2)=1-x+x^2-x^3,
$$
which is equivalent to
$$
x^3\leq x^2/2,
$$
which is true when $0<x\leq 1/2$.
Suppose $x$ is negative:
The power series expansion for $\frac{1}{1-x}$ is $\sum_{i=0}^\infty x^i$ is alternating for $-1>x>0$.  Therefore, for any even $m$,
$$
\sum_{i=0}^m x^i\geq\frac{1}{1-x}.
$$
On the other hand, when $-1<x<0$, $x^2+x<0$, so the power series expansion for $e^{x^2+x}$ is $\sum_{i=0}^{\infty}\frac{(x^2+x)^i}{i!}$ is alternating.  Therefore, for any odd $n$,
$$
\sum_{i=0}^{\infty}\frac{(x^2+x)^i}{i!}\leq e^{x^2+x}.
$$
When $m=2$ and $n=1$, we have the inequality
$$
1+x+x^2\leq 1+x+x^2,
$$
which is true for all $x$.
A: Hint :
As $f(x)=e^{x^2}$ is convex we have the inequality for $x\in(0,\varepsilon)$ with $\varepsilon>0$ sufficiently small
$$f'\left(0\right)x+f\left(0\right)+\frac{f''\left(0\right)}{2}x^2\leq f(x)$$
Where we have used strong convexity .
Now we have to show $x\in(0,\varepsilon)$ with $\varepsilon>0$ sufficiently small :
$$e^{-x}\leq \left(1-x\right)\left(f'\left(0\right)\left(x\right)+f\left(0\right)+\frac{f''\left(0\right)}{2}\left(x^{2}\right)\right)$$
Or :
$$e^{-x}\le (1-x)(1+x^2)$$
You can differentiate third times and come back to the inequality .
Ps: It's not particulary elegant but it works .
Edit : We don't need strong convexty we can use directly :
$$(1+x^2)\leq e^{x^2}$$
For $x>0$
A: Yet another approach:  Let
$$f(x) = \frac{e^{-x}}{1-x}\,;\qquad g(x)=e^{x^2}$$
then with a bit of computation, get the first and second derivatives as:
$$
f'(x) = \frac{x}{(1-x)^2} e^{-x}\,;\qquad 
g'(x) = 2xe^{x^2}
$$
$$
f''(x) = \frac{1+x^2}{(1-x)^3}e^{-x} \,;\qquad
g''(x) = (4x^2+2)e^{x^2}
$$
Now observe that $f(0) = g(0)$ and $f'(0)=g'(0)$ and $f''(0) < g''(0)$ witch implies that $f(x) \leqslant g(x)$ for all $x$ in some neighborhood of 0.  So in the remainder let's show that theorem.  It's pretty intuitive, but I don't know if it has an estabished name.  Apply the theorem to the difference $h=g-f$.

Theorem: Let $h$ be a function in $C^2(\Bbb R)$ such that $h(0) = 0$, $h'(0) = 0$ and $h''(0) > 0$.  Then there is a neighborhood $U$ of 0 such that $h(x)\geqslant 0$ for all $x\in U$.$\def\e{\varepsilon}$

To show this, there is some neighborhood $U$ of 0 and $\e>0$ such that $h''(x) \geqslant \e$ for all $x\in U$.  This is because $h''$ is continuous and $h''(0)> 0$. Integrating from 0 to $x\in U$ yields:
$$\e x = \int_0^x \!\!\e\,dt \leqslant \int_0^x h''(t)\, dt \stackrel{(1)}= h'(x) \tag3$$
for all $x\in U$, where equality $(1)$ holds because $h'(0)=0$. Integrating from $0$ to $x\in U$ again:
$$0\leqslant \frac12 \e x^2 = \int_0^x \!\!\e t\,dt \stackrel{(3)}\leqslant \int_0^x h'(t)\, dt \stackrel{(2)}= h(x)\tag 4$$
for all $x\in U$, where equality $(2)$ holds because $h(0)=0$. This concludes the proof.  It allows for a result that's a bit stronger, namely that $h(x) > 0$ for all $x$ in a punctured neighborhood of 0.
