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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!

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  • $\begingroup$ 1)$x^x=e^{x\ln x}$ , $e^x$ and $x\ln x$ both differentiable. $\endgroup$ – Vim Feb 23 '15 at 13:59
  • $\begingroup$ Also remember that $x\ln x<x^2$ $\endgroup$ – Empy2 Feb 23 '15 at 14:00
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    $\begingroup$ I actually believe reverse inequality is true - for example, for $x=1$ we have $x^x=1<e=e^{x^2}$ $\endgroup$ – Wojowu Feb 23 '15 at 14:02
  • $\begingroup$ @Wojowu edited, you right $\endgroup$ – user2637293 Feb 23 '15 at 14:07
  • $\begingroup$ @Michael why $xlnx<x^2$ ? $\endgroup$ – user2637293 Feb 23 '15 at 14:10
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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 $x>\ln x$.

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  • $\begingroup$ sorry for mistaken by me, i editd $\endgroup$ – user2637293 Feb 23 '15 at 14:07
  • $\begingroup$ for a proof of $\ln(x)<x$ see math.stackexchange.com/questions/847034/… $\endgroup$ – Surb Feb 23 '15 at 15:01
  • $\begingroup$ @Surb Just need to show $$\ln(x)<\ln(1+x)\le x$$ $\endgroup$ – Vim Feb 23 '15 at 15:12
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$e^{x^2} = (e^x)^x \geq x^x$

It is a well known fact that $e^x \geq x$, so we are done.

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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.

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For $x>0:$

$\begin{align} x &>\ln x \\ \therefore \quad x^2 &> x\ln x\\ \therefore \quad e^{x^2} &> x^x\\ \end{align}$

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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 $

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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: $(e^{x\ln(x)})'= (\ln(x) + \frac{x}{x})e^{x\ln(x)}$

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