Why $e^x$ is always greater than $x^e$?

Question

I find it very strange that $$ e^x \geq x^e \, \quad \forall x \in \mathbb{R}^+.$$

I have scratched my head for a long time, but could not find any logical reason. Can anybody explain what is the reason behind the above inequality? I know this is math community, but I'd appreciate a more intuitive explanation than a technical one.

Answer 1

Not a rigorous proof at all, but you could take the $xe$th root of both sides, and then we have

$$ e^{1/e} \geq x^{1/x} $$

That $x^{1/x}$ attains a maximum at $x = e$ can be shown fairly straightforwardly.

ETA2: A picture is worth—well, a lot of words, if not quite a thousand:

ETA: OK, a discussion of why $x^{1/x}$ attains a maximum at $x = e$. To obtain the actual value requires calculus, but we can get some intuition for the fact that it attains a maximum somewhere near $e$, as follows.

Let $f(x) = x^{1/x} = \sqrt[x]{x}$. For $x = 1$, we clearly have $f(x) = 1$. For $x = 2$, we have $f(x) = \sqrt{2} > 1$. So far, it's going up.

Now, for $x = 4$, we have $f(x) = \sqrt[4]{4} = \sqrt{2} = f(2)$. And for $x$ really large—let's say, $x = 256$, we have $f(x) = \sqrt[256]{256} = \sqrt[128]{16} = \sqrt[64]{4} = \sqrt[32]{2}$. We may not know exactly what the $32$nd root of $2$ is, but it's clearly quite a bit less than the square root of $2$. (It's about $1.02190$.) However, no matter how large $x$ is, $f(x)$ is always the $x$th root of a number greater than $1$, so $f(x)$ must be greater than $1$.

The general shape we have, then, is a function that (over the range from $1$ to $\infty$) starts at $1$, rises to a peak at some point between $2$ and $4$, and then falls off toward $1$ again (though never arriving at it). We can even find a value of $x$ in the interval $(2, 4)$ for which $f(x) > \sqrt{2}$. That is,

$$ f(\sqrt{8}) = \sqrt{8}^{1/\sqrt{8}} $$

How does this compare to $f(2) = f(4) = \sqrt{2}$? At first glance, it is difficult to say. However, if we cube $f(\sqrt{8})$, we get

$$ [f(\sqrt{8})]^3 = \sqrt{8}^{3/\sqrt{8}} $$

Note that that exponent is greater than $1$ (since $3 = \sqrt{9} > \sqrt{8}$), so

$$ [f(\sqrt{8})]^3 = \sqrt{8}^{3/\sqrt{8}} > \sqrt{8} = [f(2)]^3 = [f(4)]^3 $$

so (again, not rigorously)

$$ f(\sqrt{8}) > \sqrt{2} = f(2) = f(4) $$

The actual value is about $1.44426$.

As I said previously, proving that $x^{1/x}$ attains a maximum specifically at $x = e$ requires calculus, but it seems that may not be exactly what you want? Hoping the above serves as some kind of intuitive basis.

\begin{align} \frac{d}{dx} x^{1/x} & = \frac{d}{dx} e^{\frac{\ln x}{x}} \\ & = \left( \frac{d}{dx} \frac{\ln x}{x} \right) e^{\frac{\ln x}{x}} \\ & = \frac{1-\ln x}{x^2} e^{\frac{\ln x}{x}} \\ & = \frac{1-\ln x}{x^2} x^{1/x} \end{align}

and $f(x)$ attains a maximum when that derivative is equal to $0$, which happens when $1 = \ln x$, which happens when $x = e$. (There might be a bit of circularity there—I have to give that some thought—but that's how you obtain the maximum.)

Answer 2

Let $f:\mathbb{R}^+\to\mathbb{R}$ given by $f(x)=e^x-x^e$.

Then $f^\prime(x)=e^x-ex^{e-1}=e(e^{x-1}-x^{e-1})$ and $f^{\prime\prime}(x)=e(e^{x-1}-(e-1)x^{e-2})=e(e^{x-1}-ex^{e-2}+x^{e-2})$. Thus $f^{\prime}(e)=0$ and $f^{\prime\prime}(e)=e^{e-1}>0$. So $f$ has a local minimim in $e$.

Since $\lim_{x\to\infty}f(x)=\infty$, thus $f$ attains his absolut minimum on $x=e$, which is $0$.

Then $f(x)\ge 0$ for all $x\in\mathbb{R}^+$.

Answer 3

Here is a different approach, just for the sake of variety, which could be made more rigorous:

Consider the tangent to the curve $y=\ln x$ at the point $x=e,y=1$ is $$y-1=\frac 1e(x-e)\implies y=\frac xe$$

We know that the curve is concave so lies below the tangent except at the tangent point.

Therefore, $$\ln x\leq\frac xe$$ $$\implies x^e\leq e^x$$

Answer 4

The reasons is that the exponential function($something^x$) grows much faster than the power function ($x^{something}$)

So in general you would expect that for every $a, b > 1$something like this hold

$$a^x \ge x^b$$ at least for $x$ big enough

Now I don't think it's particularly meaningful that for the special case $a = b = e$ the inequality holds for every $x \ge 0$. It can be shown easily with basic calculus in any case, but maybe that won't satisfy your intuition :)

Answer 5

I'd appreciate a more intuitive explanation than a technical one.

Well on that, try to consider the behaviour of $2^x$. Every time I increase x by one I get another (binary) digit out of my expression. So I quickly get a huge number. If I double $x$ I get way more digits.

Now compare that with $x^2$. Every time I increment $x$ I will not generally get an extra digit from this expression. In fact as $x$ gets really large there may be no difference in the number of digits between two "adjacent" values of the expression.

For example the number of binary digits required to represent $1000^2$ is the same as the number to represent $1001^2$. In fact it's 10 binary digits for both. But $2^{1000}$ requires one thousand binary digits. So a huge difference in growth.

So we see an exponentiation function grows much faster than the square function.

If binary digits aren't your "thing" then use base $10$ and compare $10^x$ and $x^{10}$. It's the same argument.

Now if you consider your specific request about $e^x$ and $x^e$ you should see, intuitively, that this is a similar situation in terms of relative growth. We no longer have the convenience of binary digits, but it's pretty clear the same rationale holds.

Answer 6

$$e^x \ge x^e ,\quad \forall x \in \mathbb{R}^+$$ $$\ln e^x \ge \ln x^e$$ $$ x\ln e \ge e \cdot \ln x$$ $$ x \ge e \cdot \ln x$$

Let $f(x) = x-e\cdot \ln x.$ Using derivatives to find an absolute minimum, you'll find one at $x=e$, where $f(e)=0$. So at $[0, e)\cup (e, \infty)$ we have $f(x)>0 \implies x \gt e \cdot \ln x,$ and at $x=e$ we have $f(x)=0 \implies x = e \cdot \ln x.$

$$\therefore e^x \ge x^e ,\quad \forall x \in \mathbb{R}^+.$$

Answer 7

Throughout, $\log$ denotes the natural logarithm.

If $x$ and $y$ are positive real numbers, then \begin{align*} x^{y} \leq y^{x} &\quad\text{if and only if}\quad y\log x \leq x \log y \\ &\quad\text{if and only if}\quad \frac{\log x}{x} \leq \frac{\log y}{y}. \end{align*} The final inequality holds for all positive real $x$ if and only if $y$ is an absolute maximum of the function $$ \phi(u) = \frac{\log u}{u},\quad u > 0. $$ It turns out the function $\phi$ has a unique maximum, at $u = e$. This tells you two things:

• $x^{e} \leq e^{x}$ for all $x > 0$, and

• $b = e$ is the only positive real number with the property that $x^{b} \leq b^{x}$ for all $x > 0$.

Answer 8

The number $e$ is somewhat accidental, but there is a simple graphical explanation that applies to many other two-variable functions more general than $x^y$.

The function $f(x,y)=x^y$ has the property that the graph of $f(x,y)=f(y,x)$ is a transverse (in fact, perpendicular) intersection of the line $y=x$ with a curve that is the graph of a continuous function on some interval of the real line. The intersection point $x=y=E$ will have either $f(E,x) \geq f(x,E)$ for all $x$, or the opposite inequality for all $x$.

Why? The line and the curve divide the region above and below the interval into 4 parts, and in the interior of each part the difference $D = f(x,y) - f(y,x)$ has constant sign. Opposite parts have the same sign, therefore a horizontal or vertical line through $E$ have $D$ of one sign.

For the function in this question $x^y$ it is not hard to derive that $E=e$ is the intersection point. It can be done without calculus using the parameterization of $x^y=y^x$ as $x = (1+\frac{1}{w})^w$ , $y = (1 + \frac{1}{w})^{w+1}$, which at $w = \infty$ gives $x=y=e$.

Which parts of this generalize to other functions? Properties that are important for this $f$ are that the vertical cross-sections $D(a,x)$ are convex and the horizontal sections are negatives of the verticals. The graph of $D = f(x,y)-f(y,x)$ is a saddle surface with positive and negative curvatures are each point; the Hessian is of signature $(1,1)$ everywhere. I do not have a general characterization but it is possible that once a point on the diagonal is found with $(E,E)$ a local minimum of $D(x,y)$ in the vertical direction, and the saddle condition holds, then $f(E,x) - f(x,E)$ does not change sign.