How to demonstrate that #sqrt(2)^(sqrt(3)-1)>sqrt(3)^(sqrt(2)-1)#?We know that #g:(1,+oo)->RR,g(x)=##(lnx)/(x-1)#,and #g# is decreasing.

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Answer 1 · 2017-04-03T17:33:17

See below.

We have to compare

#(sqrt(a))^(sqrt(b)-1)# and #(sqrt(b))^(sqrt(a)-1)#

Applying #log# to both sides

#(sqrt b - 1)log(sqrt(a)) > (sqrta -1)log(sqrt(b))# or

#log(sqrt(a))/(sqrta - 1) > log(sqrtb)/(sqrtb - 1)#

The function #g(x) = log(x)/(x-1)# is strictly monotonic decreasing so

if #a > b# then #g(b) > g(a)#. Here #a = sqrt(2) < b = sqrt(3)# then

#g(2) > g(3)# as proposed.