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 $p le e$

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