<s>[INST] The proof of $e^x \leq x + e^{x^2}$ How can we prove the inequality $e^x \le x + e^{x^2}$ for $x\in\mathbb{R}$? 
 [/INST] 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
$$
e^{x^2} - e^x + x
> e^{x} - e^x + x > 0.
$$
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