The proof of $e^x \leq x + e^{x^2}$ How can we prove the inequality $e^x \le x + e^{x^2}$ for $x\in\mathbb{R}$?
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 $-1 &lt; x &lt; 1$, then
\begin{align*}
e^{x^2} - e^x + x
&amp;\ge x^2 + x + 1 - \sum_{k \ge 0} \frac{x^k}{k!} \\
&amp;= \frac{x^2}{2!} - \sum_{k \ge 3} \frac{x^k}{k!} \\
&amp;\ge \frac{x^2}{2!} - \sum_{k \ge 3} \frac{x^2}{k!} \\
&amp;= x^2 \left( \frac12 - \left[e - 1 - \frac{1}{1!}
- \frac{1}{2!} \right]\right) \\
&amp;= x^2 \left( 3 - e \right) \ge 0.
\end{align*}

Finally, if $x \ge 1$, then
$$
e^{x^2} - e^x + x
&gt; e^{x} - e^x + x &gt; 0.
$$
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!