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How does one prove that $$e^x \ge x^e$$ for all $x \ge 0$?

I tried to do this by setting $f(x)=e^x-x^e$

Plotting this function shows this easily, as seen here.

However, when I tried to prove this, it proved quite difficult. It seems to be increasing for $0 \le x \le e$, and seems to be increasing for all $x \ge e$.

I tried to use that $f'(x)=e^x-ex^{e-1}$ but was not able to.

Any help would be appreciated.

Travis Willse
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S.C.B.
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    Note that this is equivalent to $e^{1/e}\geq x^{1/x}$ and then show the function $x^{1/x}$ has a maximum at $x=e$. – Thomas Andrews Feb 05 '16 at 14:43
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    Many answers to the same question at http://math.stackexchange.com/questions/1619911/why-ex-is-always-greater-than-xe . – zyx Feb 05 '16 at 14:55
  • Your attempt would probably carry through if you consider the for these purposes equivalent function $f(x)=e^{\frac xe}-x$. – Lutz Lehmann Feb 05 '16 at 15:38

4 Answers4

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Hint Since $x \mapsto \log x$ is a strictly increasing function, applying it to both sides of the inequality gives that it is equivalent to $$x \geq e \log x .$$

There are at least two options here:

(1) Rearrange the inequality as $\frac{\log x}{x} \leq \frac{1}{e}$, and analyze the expression on the l.h.s. of this equality. In particular, show that its maximum value is (at most) $\frac{1}{e}$.

(2) From inspecting the original equation, we know that both sides of the inequality are equal at $x = e$. So, if the inequality holds we must have that the derivatives of both sides of the inequality agree at that point, and indeed, this is straightforward to check. These facts, together with $\frac{d^2}{dx^2} \log x < 0 = \frac{d^2}{dx^2} x$ are sufficient to give the inequality.

Travis Willse
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  • Or you could just simply say that the tangent of the concave function $\ln x$ at $(e,1)$ is always above the function, which implies that $$ \ln x \le 1+\frac{x-e}{e}=\frac{x}{e}.$$ – Lutz Lehmann Feb 05 '16 at 15:35
  • This is true, but saying "tangent of the concave function... is always above..." is essentially inequivalent to (2), no? – Travis Willse Feb 05 '16 at 15:37
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If $x=0$, then given inequality holds. Suppose that $x > 0$ and define a function $f(x)=x - e\ln x$. The derivative of $f(x)$ is $f'(x)=1-\frac{e}{x}$. Thus $f(x)$ decreases when $x<e$ and increases when $x > e$. $f(e)=e-e\ln e =0$ and $f(e)$ is minimum of $f(x)$, so $f(x)\ge 0$ for all $x > 0$. Thus, $x\ln e \ge e\ln x$ for all $x > 0$ and so $\ln e^x \ge \ln x^e$. Since $\ln x$ strictly increases, $e^x \le x^e$ for all $x > 0$. We previously showed that the inequality holds at $x=0$, so we get the conclusion expected.

choco_addicted
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The graph $\gamma$ of $\log$ is concave, hence $\gamma$ stays below the tangent to $\gamma$ at $(e,\log e)$: $$\log x\leq \log e+{1\over e}(x-e)\qquad(x>0)\ .$$ This simplifies to $\log x\leq{x\over e}$, and exponentiation gives $$x^e\leq e^x\qquad(x>0)\ .$$ The case $x=0$ is obvious.

Christian Blatter
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Consider the function $$f(x)=e^x-x^e$$ The derivative is $$e(e^{x-1}-x^{e-1})$$ It is clear that $$f'(1)=f'(e)=0$$ and it is verify that $x=1$ goes for a maximun and $x=e$ goes for a minimun. Furthermore $f(1)=e-1>0$ and $f(e)=e^e-e^e=0$.

Taking in account that $f(0)=1$ and the domain of $f$ is $x\ge 0$ it follows the conclusion $$e^x\ge x^e$$

Piquito
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