
derivative inequality
:mad:can someone help me??!
I have this problem..
if $\displaystyle x(t)^{'} $ is a nonnegative and decreasing function and
$\displaystyle x(t) $ is a positive and nondecreasing
so..
$\displaystyle \frac{x^{'}(t)}{x(t)}\leqslant \frac{1}{t}$ for all $\displaystyle t\geqslant b $ ??!
I tried to prove this problem but at this step
$\displaystyle x(t)x(b)\geqslant(tb)x(t)^{'} $
I don't know how to get the rest of proof from this step..:confused:
can someone help me to get the rest of proof??
thanks for any help

What is $\displaystyle x(t)$, what is $\displaystyle x'(t)$, on what interval are they defined? What is $\displaystyle b$?

moredetial
if$\displaystyle x^{'}(t) $ is a nonnegative and decreasing function
and $\displaystyle x(t) $ is a positive and nondecreasing on [b,t]
then
$\displaystyle x(t)x(b)=\int^{t}_{b}{x^{'}(s)ds} $
since $\displaystyle x^{'}(t) $ is decreasing
$\displaystyle x^{'}(t)\leqslant x^{'}(s) $
this implies
$\displaystyle x(t)x(b)\geqslant x^{'}(t) \int^{t}_{b}{ds} $
so
$\displaystyle x(t)x(b)\geqslant(tb)x^{'}(t) $
i.e.
$\displaystyle \frac{x^{'}(t)}{x(t)}\leqslant \frac{1}{tb} $.....(*)
now,I just want to get
$\displaystyle \frac{x^{'}(t)}{x(t)}\leqslant \frac{1}{t} $??!
I hope that's clear