Solution:
(I) Proof: From $a_1=1$, $a_{n+1}= \dfrac {a_n}{a_n^2+1}$, we get $a_n > 0$ for all $n\in\mathbb{N}$,
then $a_{n+1}-a_n= \dfrac {a_n}{a_n^2+1}-a_n= \dfrac {- a_n^3}{ a_n^2+1} \dfrac {1}{2}a_n$,
$\therefore a_n > \dfrac {1}{2}a_{n-1}\geqslant ( \dfrac {1}{2})^2a_{n-2}\geqslant …\geqslant ( \dfrac {1}{2})^{n-1}a_1= \dfrac {1}{2^{n-1}}$, i.e., $a_n\geqslant \dfrac {1}{2^{n-1}}$.
From $a_{n+1}= \dfrac {a_n}{a_n^2+1}$, then $\dfrac {1}{a_{n+1}}=a_n+ \dfrac {1}{a_n}$,
$\therefore \dfrac {1}{a_{n+1}}- \dfrac {1}{a_n}=a_n$,
$\therefore \dfrac {1}{a_2}- \dfrac {1}{a_1}=a_1=1$, $\dfrac {1}{a_3}- \dfrac {1}{a_2}=a_2= \dfrac {1}{2}$, $\dfrac {1}{a_4}- \dfrac {1}{a_3}=a_3=( \dfrac {1}{2})^2… \dfrac {1}{a_n}- \dfrac {1}{a_{n-1}}=a_{n-1}\geqslant ( \dfrac {1}{2})^{n-2}$,
Summing up, we get $\dfrac {1}{a_n}- \dfrac {1}{a_1}=1+ \dfrac {1}{2}+( \dfrac {1}{2})^2+…+( \dfrac {1}{2})^{n-2}= \dfrac {1- \dfrac {1}{2^{n-1}}}{1- \dfrac {1}{2}}=2-( \dfrac {1}{2})^{n-2}$,
and since $a_1=1$,
$\therefore \dfrac {1}{a_n}\geqslant 3-( \dfrac {1}{2})^{n-2}= \dfrac {3\cdot 2^{n-2}-1}{2^{n-2}}= \dfrac {3\cdot 2^{n}-4}{2^{n}}$,
$\therefore a_n\leqslant \dfrac {2^{n}}{3\cdot 2^{n}-4}$.
In conclusion, we have $\boxed{\dfrac {1}{2^{n-1}}\leqslant a_n\leqslant \dfrac {2^{n}}{3\cdot 2^{n}-4}}$.