Prove that $e^{x^2}\ge x^x$ for $x\ge0$ How to prove that  $e^{x^2}\ge x^x$  for $x\ge0$?
I thought about declare a new funtion $f:\mathbb{R}^+\rightarrow \mathbb{R}$ like this:
$$f\left(x\right)\:=\:x^x-\:e^{x^2}\:$$.
1) First, i need to show that f is differentiable at $\mathbb{R}^+$, But how i can show that $\:x^x$ is differentiable there?
2)Suppose that i showed that f is differentiable at $\mathbb{R}^+$. Now, i get this:
$$f'\left(x\right)\:=\:-2\:e^{\left(x^2\right)}\:x+x^x\:\left(1+log\left(x\right)\right)$$
how to show that is $>0$?
Thanks!
 A: I think you better use this alternative
$$x^x> e^{x^2}\Leftrightarrow\ln(x^x)>\ln(e^{x^2})\Leftrightarrow x\ln x>x^2\ln e\Leftrightarrow \ln x>x$$
For all $x>0$. But I guess it should be "$<$" rather than "$>$" in what you want to prove, as the final result is actually $x>\ln x$.
A: $e^{x^2} = (e^x)^x \geq x^x$
It is a well known fact that $e^x \geq x$, so we are done.
A: For all $x>0$, we have $\ln x\le x$. So $x\ln x\le x^2$, which implies $e^{x\ln x}\le e^{x^2}$. Finally, notice that $e^{x\ln x}=x^{x}$, we get the conclusion.
A: 1) $x^x=e^{x\ln x}$. Now you can Show that $x\ln x$ is differentiable by considering difference quotients.
2) Use the chain rule. Hint: $(e^{x\ln(x)})'= (\ln(x) + \frac{x}{x})e^{x\ln(x)}$
A: If your goal is to show that $ e^{x^2} ≥ x^x $ then why not do this?
$ f(x) = e^{x^2} - x^x $
Note:
$ f(0^+) = 0 $
Observe that 
$
\lim_{x \to \infty} \frac {e^{x^2}}{x^x} =
\lim_{x \to \infty} \frac { e^{x^2}}{e^{x\ln x}} \\ =
\exp(\lim_{x \to \infty} \frac {x^2}{x\ln x}) \\ =
\exp(\lim_{x \to \infty} \frac {x}{\ln x}) = \exp(\infty) = \infty 
$
(I'd say this would be sufficient, but...)
We can also show:
$
\lim_{x \to \infty} \frac {x^x(\ln x+1)}{2xe^{x^2}} \\ \ =
0.5\lim_{x \to \infty} \frac {x^{x-1}\ln x+x^{x-1}}{ e^{x^2}} \\ \ =
0.5\lim_{x \to \infty} (\frac {x^{x-1}\ln x}{ e^{x^2}} + \frac {x^{x-1}}{ e^{x^2}}) \\ \ =
0.5(0 + 0) = 1 \text{ (using what we shown before)}
$ 
So now we know that $ 2xe^{x^2} $ increases to $\infty$ faster than $ x^x(\ln x+1) $, hence
$
\frac {df} {dx} = 2xe^{x^2} - x^x(\ln x+1) ≥ 0
$
(please note that by showing $\frac d {dx} x^x$ exist for all x, and $\frac d {dx} e^{x^2}$ for all x, you have shown that f(x) has a derivative for all x.
so since $f(0^+)=0$ and the function is increasing we conclude that on $ (0,\infty) $ the function is $f(x)≥0$,
and thus $ e^{x^2} ≥ x^x $
A: For $x>0:$
$\begin{align}
 x &>\ln x \\
\therefore \quad x^2 &> x\ln x\\
\therefore \quad e^{x^2} &> x^x\\
\end{align}$
