The proof of $e^x \leq x + e^{x^2}$ How can we prove the inequality $e^x \le x + e^{x^2}$ for $x\in\mathbb{R}$? 
 A: Note the inequality $e^t \ge t + 1$ for all $t \in \mathbb{R}$.
In particular $e^{x^2} \ge x^2 + 1$.
If $x \le -1$, then
$$
e^{x^2} - e^x + x \ge x^2 + 1  - e^0 + x = x(x+1) \ge 0.
$$
If $-1 < x < 1$, then
\begin{align*}
e^{x^2} - e^x + x
&\ge x^2 + x + 1 - \sum_{k \ge 0} \frac{x^k}{k!} \\
&= \frac{x^2}{2!} - \sum_{k \ge 3} \frac{x^k}{k!} \\
&\ge \frac{x^2}{2!} - \sum_{k \ge 3} \frac{x^2}{k!} \\
&= x^2 \left( \frac12 - \left[e - 1 - \frac{1}{1!}
- \frac{1}{2!} \right]\right) \\
&= x^2 \left( 3 - e \right) \ge 0.
\end{align*}
Finally, if $x \ge 1$, then
$$
e^{x^2} - e^x + x
> e^{x} - e^x + x > 0.
$$
A: You have that $e^1< 1+e^{1^2}$. Now see what happens with the derivative on each side, e.g., we have for $x>1$ that $e^x < 1+2xe^{x^2}$ , since $2x>1$ and $e^{x^2}>e^{1}=e$ , so that , when $x>1$, the left-hand side is smaller than the R.H side, and then the R.H side grows faster nafterwards, for $x>1$. Try seeing what happens left of $1$.
We also have that for $x=1 , e^{-1}<-1+e^1$ , and $h(x)=1+2xe^{x^2}$ decreases faster than $e^{x}$ in $(-\infty,-1]$ (Meaning $h(|x|) $ grows faster than $e^{|x|}$). I don't think this is too hard to prove; for $x<-1$ , we have $1-2e^{x^2}< e^{x}$.
The whole issue comes down to the fact that $e^{x^2}$ grows way faster than $e^x$ in $[1, \infty)$ and decreases way faster in $(-\infty, -1]$. In $[-1,1]$ the result still holds; it is a little late now, but I will have a proof by Tuesday night, or I will delete this.
A: $$e^{x} = 1 + x + \frac{x^2}{2!} + \dots \le 1 + x + 2 \left(\left( \frac{x}{2} \right )^2+\left( \frac{x}{2} \right )^3 +\dots \right)\le 1 + x+\left( \frac{x}{2} \right )^2 \left( \frac {4}{2 - x}\right)$$
$$x + e^{x^2} \ge  x +1  + x^2 + \frac{x^4}{2!} = 1+ x+  \left( \frac{x}{2} \right )^2 2(2+x^2) \ge 1+ x+  4  \left( \frac{x}{2} \right )^2 $$
The inequality is true for all $\Bbb R \setminus (1,3)$, let $b \in (1,2)$, 
Also,
$$e^{(1+b)^2}+1+b \ge  e^{(1+b)^2} \ge e^{1+b}$$
hence the inequality holds for all $x\in \Bbb R$
A: Hmmm.... I will give it a try. Consider the function $f(x) = e^{x²} + x - e^x$, $x \in \mathbb{R} $. Find the global minimum of this function. It turns out to be zero. Does that prove it ?
Note: The second derivative is $e^{x^2} (4 x^2+2)-e^x$ which is strictly positive.  $e^{x^2}$ becomes less than $x$ between $0$ and $1$ but that difficulty is overcome by multiplying by 2 (*)
(*)That difficulty is not overcome just by multiplying by 2, I learned it the hard way. You still have to prove it !
I did not answer the question, but I will keep the post here. Reminding me of my once failed efforts!
