Show that $e^x\ge x^e$ for all $x\ge 0$ Show that $e^x\ge x^e$ for all $x\ge0.$
I was playing around with splitting a number into parts (that add to the original number) and then multiplying the parts to get the result, and I found that the product is highest when the parts tended towards $e$. This inequality is a consequence of that. Is there a proof or intuitive explanation for this?
 A: It suffices to prove for all $x\neq 0$ $$f(x)=x-e\log x\ge 0$$ $$f'(x)=0\to 1-\frac{e}{x}=0\to x=e$$ also  here`$f''(e)>0$ so $$f(x)\ge f(e)=0$$ note that $$\lim_{x\to 0^+}f(x),\lim_{x\to \infty} f(x)>0$$
A: Start with $e^x \geq 1 + x$. Substitute $x \leftarrow x - 1$ to find $e^x \geq e x$. Substitute $x \leftarrow x/e$ to find $e^{x/e} \geq x$.
A: Outline:  At $x = 0$, $e^x = 1 > 0 =x^e$ and and at $x = e$ we have $e^x = x^e$. To determine if they are equal anywhere else or if $x^e$ is ever greater than $e^x$ we can consider the rates of change (i.e. the derivatives)
But it will be easier to consider the the $\log e^x = x$ and the $\log x^e= e\log x$ because their derivatives ($1$ and $\frac ex$) are really clear and apparent.  We can do  this safely as $\log y$ is montotonically increasing so $0 < a < b \iff \log a < \log b\iff e^a < e^b$.
So $e^x \ge x^e \iff x \ge e \ln x$.  And $x = 0$ we have $x > \log 0$ which... well admitted $\log 0$ is undefined but $\lim_x\to 0 \log x = -\infty$ so it is kosher to so $x = 0> \log 0$.
At $x=e$ the $x =e = e\ln x$ but $\frac {dx}{dx}= 1$ and $\frac {de\ln x}{dx} = \frac ex$. And $x =e$ these rates are equal but for any $x > e$ then $1 > \frac ex$ so $x$ is increasing faster than $e\ln x$ and so they can never be equal for any point $x > e$.
And for $0 < x < e$ then $\frac {dx}{dx} = 1 < \frac ex =\frac {dx \ln x}{dx}$  so on $(0,e)$ we have $e\ln x$ is increasing faster than $x$.  So if there is any $x\in (0,e)$ so that $x \le e\ln x$ then $e\ln x$ would surpass $x$ and $x$ would never be able to catch up until $\frac {dx}{dx}>\frac {de\ln x} {dx}$ but that never happens in $(0,e)$.
So it is not possible for $x \in (0,e)$ for $x \le x\ln x$.
So $x > e\ln x$ for all $x < e$ and for all $x > e$ and at $x =e$ we have $x =e\ln e$.
So $x \le e\ln x$ for all $x \ge 0$.
And so $e^x \le e^{e\ln x} = x^e$ for all $x \ge 0$.
A: Having that $\dfrac{\log x}{x}$ is a increasing function with $x\leq e$
$$\therefore\frac{\log e}{e}\geq\frac{\log x}{x}$$
Excluding the logarithm signs in both side, we have what we need to show.
