I stumbled upon this problem and can't find a solution. I tried taking the derivative of this function $e^x - x^e$ and got this function: $e(e^{x-1} - x^{e-1})$ and I can't prove it is positive or equal to zero because it's almost my original problem. If i could prove the derivative was always positive then i could say that the function is strictly increasing and because the limit as $x \to 0^{+}$, of the function is positive then it must mean that $e^x \ge x^e$.
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5Hint: take logs of both sides to see that this is the same as $x≥e\ln ,x$. Now, where does the function $f(x)=x-e\ln ,x$ have a minimum? – lulu Oct 23 '16 at 18:08
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1Hint: $x^e = \exp(e \cdot \log(x))$ – user251257 Oct 23 '16 at 18:09
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3Duplicate: Why $e^x$ is always greater than $x^e$? – Workaholic Oct 23 '16 at 18:10
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The minimum value will be $f(e)$ which is 0 so $f(x) >= 0$, $x>0$ – Ciocan Cosmin Oct 23 '16 at 18:20
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You're better to consider the function $$ f(x)=x-e\log x $$ (natural logarithm). The derivative is $$ f'(x)=1-\frac{e}{x}=\frac{x-e}{x} $$
egreg
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From $\log(x) \leq x-1$ for all $x>0$ it follows that $$\log(e x) = 1 + \log(x) \leq x.$$ Substitute $x \leftarrow e^{-1}x$ to get $$\log(x) \leq e^{-1}x$$ and this is an equivalent form of your inequality.
WimC
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