A Basic Inequality I want to show that $e^x \leq x + e^{x^2}$ for all $x \in \mathbb{R}$.  One can see that if $x \geq 1$ then $e^{x^2}$ dominates $e^x$.  A Taylor expansion argument of the derivative followed by term by term comparison handles the rest, but I'm curious if there is a more unified way to prove this inequality without resorting to cases.
 A: You can check that $f(x)= x e^{-x}+e^{x^2-x}$ is a convex function and notice that $f'(0)=0$.
This immediately implies that $f(x)\geq f(0)=1$ for any $x\in\mathbb{R}$.
A: By the Jack's beautiful hint we obtain.
Let $f(x)=e^{x^2}+x-e^x.$
Thus, $$f'(x)=2xe^{x^2}+1-e^x$$ and
$$f''(x)=4x^2e^{x^2}+2e^{x^2}-e^x=e^x\left(2e^{x^2-x}(2x^2+1)-1\right)\geq e^x\left(2e^{-\frac{1}{4}}-1\right)>0,$$
which says that $f$ is a convex function.
Thus, if there is a value of $x$ for which $f'(x)=0$ then $f$ gets for this value a minimal value. 
But $f'(0)=0$, which says that $$f(x)\geq f(0)=0$$
and we are done!
A: We can look at the zeroes of the function $e^{x^2}-e^x+x$, and see where their critical points are. Taking the derivative of $e^{x^2}-e^x+x$, we get
$e^{x^2}*2x - e^x + 1$ from the chain rule.
We can notice that the expression is equal to $0$ when $x = 0$. Since when $x$ is negative, $e^{x^2} * x < e^x$, and when $x$ is positive, $e^x < e^{x^2} * x$, then we can conclude that $0$ is the only turning point. Furthermore, when we substitute $x=-1$ and $x=1$ into the original function $e^{x^2}-e^x+x$, we get $e-e^{-1}-1$ and $e^1-e^1+1$, which are both positive. 
Therefore, the graph of $e^{x^2}-e^x+x$ is concave up. What can you conclude from this?
