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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?

Sarah Pierson
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  • For $x > 0$, you can rearrange this inequality to become $e^{1/e} \ge x^{1/x}$, and then $(1/e)^{1/e} \le (1/x)^{1/x}$. This inequality follows from the fact that $x^x$ has a minimum at $x = 1/e$. – Theo Bendit Mar 10 '21 at 02:31
  • Also: https://math.stackexchange.com/q/1641851/42969, https://math.stackexchange.com/q/1981731/42969, https://math.stackexchange.com/q/619119/42969 – all found with Approach0 – Martin R Mar 10 '21 at 07:49
  • Another one: https://math.stackexchange.com/q/1344288/42969. – Martin R Mar 10 '21 at 07:58
  • "intuitive explanation " Well the most intuitive explanation is that functions of the form $b^x$ (assuming $b> 1$) tend to grow faster and "outstrip" functions of the type $x^m$. In a single step $M=b^x$ jumps to $b^{x+1}=bM$ and $b^{x+1}-b^x=(b-1)M$ and that's a multiple of itself. But $K=x^m$ then $(x+1)^m-x^m= mx^{m-1}+...\approx \alpha K^{\frac {m}{m-1}}$ is only a multiple of a root of itself. SO eventually $b^x$ will always outstrip $x^m$. And $e^x$ outstripped $x^e$ from the very get go. – fleablood Mar 10 '21 at 16:47

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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 $$\lim_{x\to 0^+}f(x),\lim_{x\to \infty} f(x)>0$$

Albus Dumbledore
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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 $e^{x/e} \geq x$.

WimC
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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 $x \ge 0$.

fleablood
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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.