How can we prove that for $b>a>e$ ($e$ being the Euler’s number), $a^b$ is greater than $b^a$?
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You might have a look at Fastest way to check if $x^y > y^x$? and other questions linked there. – Martin Sleziak Mar 24 '19 at 12:14
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2Possible duplicate of Fastest way to check if $x^y > y^x$? – Martin Sleziak Mar 24 '19 at 12:15
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You want to prove $x^{1/x}$, or equivalently its logarithm $x^{-1}\ln x$, decreases in $x\ge e$ so that $a^{1/a}>b^{1/b}$. Use the product rule.
J.G.
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