Upper bound for $e^x$ 
How do I prove that $$e^x \leq x + e^{x^2}$$ for all $x\in\mathbb R$? 

My probability book (Grimmett and Stirzaker) says that it's a simple exercise but I don't see it. For $x\leq 0$, we have
$$e^x = \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + x + \sum_{k=1}^\infty \frac{x^{2k+1}}{(2k+1)!} \leq \sum_{k=0}^\infty \frac{x^{2k}}{k!} + x = e^{x^2} + x.$$
How do I show it for $x>0$?
 A: The case $x\ge1$ is trivial. For $0<x<1$, we have
$$
\frac{e^x-e^{x^2}}{x-x^2}\le e^x\le\frac1{1-x}.
$$
The first inequality is true because the LHS, by mean value theorem, is $e^a$ for some $a\in(x^2,x)$. For the second one, note that $f(x):=(1-x)e^x\le1=f(0^+)$ because $f$ is decreasing ($f'(x)=-xe^x<0$) on $(0,1)$.
A: The function
$$f(x):=e^{x^2}-e^x$$
has $f(0)=0$, $f'(0)=-1$ and
$$f''(x)=(4x^2+2)e^{x^2}-e^x\ .$$
Obviously $f''(x)>0$ when $x\leq0$ or $x\geq1$.  For $0\leq x\leq1$
note that
$$4x^2+2=\left(2x-{1\over2}\right)^2+{7\over4}+2x>1+(e-1)x\geq e^x\ ,$$
since $e-1<2$ and $\exp$ is convex. It follows that $f$ is convex, and this implies that the line $y=-x$ supports the graph of $f$ at $(0,0)$, i.e.
$$-x\leq e^{x^2}-e^x\qquad(-\infty<x<\infty)\ .$$
A: How about this:
For $x\geq 0$, we have
$$ x e^{-x}+e^{x^2-x}\geq x(1-x)+1+x^2-x=1$$
where we used the inequality $e^{u} \geq 1+u$ for all $u$.
Multiplying on both sides  by $e^x$, we find 
$$x+e^{x^2}\geq e^x$$
For $x<0$, note that $e^{x^2}\geq 1+x^2$, then
$$
x+e^{x^2}\geq1+x+x^2=3/4+(x+1/2)^2>0
$$
Consequently, we also have
$$
\left(x+e^{x^2}\right)e^{-x}\geq(x+1+x^2)(1-x)=1-x^3,
$$
which implies that 
$$
x+e^{x^2}\geq(1-x^3)e^x>e^x
$$
A: We have that


*

*for $x\ge1$
$$x\le x^2 \implies e^x\le e^{x^2}\le x+ e^{x^2}$$


*

*for $0<x<1$
$$e^{x^2}+x\ge 1+x+x^2 \stackrel{?}\ge e^x =1+x+\frac12x^2+\frac16x^3+\ldots$$
and
$$1+x+x^2 \stackrel{?}\ge 1+x+\frac12x^2+\frac16x^3+\ldots \iff \frac12x^2 \stackrel{?}\ge \frac16x^3+\frac1{24}x^4+\ldots \iff \frac12 \stackrel{?}\ge \frac16x+\frac1{24}x^2+\ldots$$
then to prove $e^x\le x+ e^{x^2}$ it suffices to show that the latter holds for $x=1$ that is
$$\frac12 \ge \frac16+\frac1{24}+\ldots \ge \frac16x+\frac1{24}x^2+\ldots$$
$$\iff 1 \ge \frac13+\frac1{12}+\frac1{60}\ldots=2\sum_{k=3}^\infty \frac{1}{k!}=2e-5$$
