Show that ${{2m} \choose {m}} \leq \frac{2^{2m}}{\sqrt{2m}}$.

Show the stronger inequality $${{2m} \choose {m}} <\frac{2^{2m}}{\sqrt{2m+1}}.$$ It is true for $m=1$. Induction step: for $m\geq 1$, $$\frac{(2m+2)!}{2^{2m+2}((m+1)!)^2}=\frac{(2m)!}{2^{2m}(m!)^2}\cdot \frac{2m+1}{2m+2}< \frac{1}{\sqrt{2m+1}}\cdot \frac{2m+1}{2m+2}\stackrel{?}{\leq}\frac{1}{\sqrt{2m+3}}.$$ So it remains to prove that $$4m^2+8m+3=(\sqrt{2m+1}\sqrt{2m+3})^2\leq (2m+2)^2=4m^2+8m+4$$ which holds.

Now, $$\tfrac{3\cdot5\cdot...\cdot(2m-1)}{2\cdot4\cdot...\cdot2m}=\sqrt{\tfrac{3(3\cdot5)(5\cdot7)...((2m-3)(2m-1))(2m-1)}{4\cdot4^2\cdot6^2\cdot...\cdot(2m-2)^2\cdot4m^2}}<\sqrt{\frac{3(2m-1)}{16m^2}}<\frac{1}{\sqrt{2m}}.$$