Problem Statment:-
Prove:-$$\dfrac{4^m}{2\sqrt{m}}\le\binom{2m}{m}\le\dfrac{4^m}{\sqrt{2m+1}}$$
My Attempt:-
We start with $\binom{2m}{m}$ (well that was obvious), to get
$$\binom{2m}{m}=\dfrac{2^m(2m-1)!!}{m!}$$
Now, since $2^m\cdot(2m-1)!!\lt2^m\cdot2^m\cdot m!\implies \dfrac{2^m\cdot(2m-1)!!}{m!}\lt 4^m$
$$\therefore \binom{2m}{m}=\dfrac{2^m(2m-1)!!}{m!}\lt4^m$$
Also, $$2^m\cdot(2m-1)!!\gt2^m\cdot(2m-2)!!\implies 2^m(2m-1)!!\gt2^m\cdot2^{m-1}\cdot(m-1)!\\ \implies \dfrac{2^m\cdot(2m-1)!!}{m!}\gt\dfrac{4^m}{2m}$$
So, all I got to was $$\dfrac{4^m}{2m}\lt\binom{2m}{m}\lt4^m$$
So, if anyone can suggest me some modifications to my proof to arrive at the final result, or just post a whole different non-induction based proof.