<s>[INST] How to show that $e^x\le x + e^{x^2}$ for any real $x $ 
Show that $e^x\le x + e^{x^2}$ for any real $x$.

I have trouble proving this inequality for $x \in (0, 1)$. Any hint?
 [/INST] Let $f(x)=x+e^{x^2}-e^x$, $f(0)=0$
$f'(x)=1+2xe^{x^2}-e^x$, $f'(0)=0$
$f''(x)=2e^{x^2}+4x^2e^{x^2}-e^x=e^x(2e^{x^2-x}-1)+4x^2e^{x^2}$
$e^{x^2-x}=e^{(x-\frac{1}{2})^2-\frac{1}{4}}\ge e^{-\frac{1}{4}}>\frac{1}{2}$, so $f''(x)>0$ for $x\in(0,1)$, done.
Another try:
By Taylor expansion about $x=0$, $e^x=1+x+\frac{x^2}{2}+R_3(x)$
$R_3(x)=\frac{e^c}{3!}x^3\le \frac{e}{6}x^3$
$x+e^{x^2}=x+(1+x^2+\cdots)>1+x+x^2$
$\frac{1}{2}>\frac{e}{6}\ge \frac{e}{6}x \Rightarrow \frac{1}{2}x^2>\frac{e}{6}x^3 \Rightarrow 1+x+x^2>1+x+\frac{x^2}{2}+R_3(x)\Rightarrow x+e^{x^2}>e^x$
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