Prove that $e^x \ge x^e$ for $x \gt 0$
I applied the natural logarithm to simplify the function and I get $$\frac{x}{\ln x}\ge e$$
How to solve these types of problems?
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1How did you get this equality from the starting inequality ? – Tom-Tom Jun 30 '15 at 08:36
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1One ought to mention in any proof that the equivalence of $e^x \geq x^e$ and the inequality $x \geq e \log x$ are equivalent because $\log$ is an increasing function---this argument does not hold if we replace it with a general function. – Travis Willse Jun 30 '15 at 08:40
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Ugh, formatting issues, I didn't get an equality at the end. It's still an inequality, I edited now. – MikhaelM Jun 30 '15 at 08:45
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1@MikhaelM Note you cannot divide by $\log x$ unless you are sure it is positive, i.e. $x > 1$. For $x \in (0, 1)$, that division will reverse the inequality. – Macavity Jun 30 '15 at 09:09
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See also: Why $e^x$ is always greater than $x^e$? and Prove the inequality $e^x \geq x^e$ for $x > 0$. – Martin Sleziak Jan 07 '22 at 08:18
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You are alsmot there.
Study the function $f(x)=\dfrac{\ln x}{x}$
Then $f'(x)=\dfrac{1}{x^2}(1-\ln x)$
$f'(x)>0$ for $x<e$ and $f'(x)<0$ for $x>e$
But $f(e)=\dfrac 1e$...
Martigan
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Oh I get it, so the function x/lnx has a minimum in e, so that f(e) = e is the minimal value that the function can take. Thus, the function x/lnx has to be greater or equal to e. Thanks mate! – MikhaelM Jun 30 '15 at 09:00
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$x \geq e\ln x \iff x - e\ln x \geq 0$. $f(x) = x - e\ln x \to f'(x) = 1-\dfrac{e}{x}=\dfrac{x-e}{x}\geq 0 \iff x \geq e$. Thus if $x \geq e, f(x) \geq f(e) = 0 \to x \geq e\ln x \to e^x \geq x^e$, and if $x \leq 0, f(x) \geq f(e) = 0 \to e^x \geq x^e$.
DeepSea
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You could define the function \begin{equation}f(x)=e^x-x^e \end{equation} and show that $f'(x)$ is zero at $x=1,e$ and go forth in terms of the minima of $f(x)$. With this, however, you will also have to show that $f(x)$ doesn't achieve a minimum at $\pm\infty$...