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)#?

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Answer 1 · 2017-04-29T17:14:17

#p le e#

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#