<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] If $x \geq 1$, then $x^2 \geq x$, we then have $e^x \leq e^{x^2}$, which in turn means that $e^x \leq e^{x^2}+x$.
For $0 \leq x \leq 1$, then $x^2 \leq x$, we then have $e^x$, we then have
\begin{align}
e^x & = \sum_{k=0}^{\infty} \dfrac{x^k}{k!} = 1 + x + \sum_{k=2}^{\infty} \dfrac{x^k}{k!} \leq 1 + x + \sum_{k=2}^{\infty} \dfrac{x^2}{2^{k-1}} \,\,\,\, (\because x \in [0,1] \text{ and }k! \geq 2^{k-1})\\
& = 1+x+x^2 \leq x + e^{x^2}
\end{align}
If $x \leq 0$, then $e^{x^2}+x \geq 1+x+\dfrac{x^2}2 = \dfrac{1+(1+x)^2}2 \geq e^x$.
Hence, $e^x \leq e^{x^2}+x$ for all $x \in \mathbb{R}$.
</s>