Having function f:(0,oo)->RR,f(x)=xlnp-plnx,p inRR;p>0,how to find p>0 sush that f(x)>=0,forall x in(0,oo)?

1 Answer
Apr 29, 2017

p le e

Explanation:

From

f(x) = xlogp-plogx if f(x) ge 0 then

logp/p ge logx/x
or

e^(logp/p) ge e^(log x/x) because e^alpha is a strictly increasing transformation

or

x^(1/x) le p^(1/p)

Calculating the maximum of y = x^(1/x) we get

(dy)/(dx)=-x^(1/x-2) ( Log_ex-1) = 0
(d^2y)/(dx^2) = x^(1/x-4) (1 - 3 x + Logx (2 x + Logx-2))

and for log_ex=1->x=e we have a global maximum for y

(d^2y)/(dx^2) =-e^(1/e-3) < 0 qualifying a maximum.

so

x^(1/x) le e^(1/e)

and then p le e