Assuming that $e<y<x$, prove that $ x^y < y^x$.
I think this must be easy, but I can't work it out. Thanks in advance for any kind of help.
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$$x^y<y^x\iff x^{1/x}<y^{1/y}$$ See http://math.stackexchange.com/questions/116112/find-the-maximum-of-fx-x1-x – lab bhattacharjee Mar 20 '16 at 13:51
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HINT
take logs to compare $x \ln y ? y \ln x$ or equivalently $\frac{x}{\ln x} ? \frac{y}{\ln y}$ (sign does not change since $\ln x > \ln y > 1$) and look at the function $x/\ln x$ to see if it's increasing or decreasing for $x > e$
gt6989b
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Hint:
This is equivalent to $y\ln x<x\ln y$, i.e. since $x, y>0$, to $$\frac{\ln x}x<\frac{\ln y}y.$$
Set $f(x)=\dfrac{\ln x}x$ and determine the variations of $f$.
Bernard
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