<s>[INST] Proof of the inequality $e^x\le e^{x^2} + x$ The question is to prove the inequality $e^x\le e^{x^2} + x$. I tried the Taylor expansion like ${e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ...$ and $x + {e^{{x^2}}} = 1 + x + \frac{{{x^4}}}{{2!}} + \frac{{{x^6}}}{{3!}} + ...$ but cannot see anything useful out of this. Anyone can provide some help? Thank you.
 [/INST] Note that for any $x$, we have
$$
(e^{x^2} + x) - e^x = \\
\frac{1}{2!}x^2 - \frac{1}{3!}x^3 + \left(\frac{1}{2!} - \frac{1}{4!}\right)x^4 - \frac 1{5!}x^5 + 
\left( \frac{1}{3!} - \frac{1}{6!} \right)x^6 -  \cdots \geq \\
\frac{1}{2!}x^2 - \frac{1}{3!}x^3 + \frac {1}{4!}x^4 - \frac 1{5!}x^5 +
\frac{1}{6!}x^6 -  \cdots =\\
e^{-x} + x - 1
$$
Consider the function $f(x) = e^{-x} + x - 1$.  We note that
$$
f'(x) = 1 - e^{-x}
$$
so that $f$ achieves its minimum at $x = 0$, where we find $f(0) = 0$.  So, $f(x) \geq 0$.
Thus, we have
$$
(e^{x^2} + x) - e^x \geq f(x) \geq 0
$$
The conclusion follows.
</s>