Question

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)`?

Answer 1

`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`