How to prove $$\frac{1}{4^n}\binom{2n}{n}\leq \frac{1}{\sqrt{\pi n}} ?$$
What i tried:
$$\frac{1}{4^n}\frac{(2n)!}{n!\times n!} = \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \frac{2n-1}{2n}.$$
Arithmetic inequality
$$\frac{2r-1+2r+1}{2}\geq \sqrt{(2r-1)(2r+1)}$$
$$2r\geq \sqrt{(2r-1)(2r+1)}$$
$$\frac{1}{2r}\leq \sqrt{\frac{1}{(2r-1)(2r+1)}}$$
$$\prod^{n}_{r=1}\frac{2r-1}{2r}\leq \prod^{n}_{r=1}\sqrt{\frac{2r-1}{2r+1}}=\frac{1}{\sqrt{2n+1}}\;\;, n\geq 1,$$
but how do I prove my original inequality for natural numbers? Help me, please!
May be not the way you want, but let $$a_n=\frac{1}{4^n}\binom{2n}{n}=\frac{1}{4^n}\frac{(2n)!}{n!\times n!}$$ Take logarithms $$\log(a_n)=\log((2n)!)-2\log(n!)-n\log(4)$$ Use Stirling approximation for the factorials and continue with Taylor to get $$\log(a_n)=-\frac{1}{2} \log \left({\pi n}\right)-\frac{1}{8 n}+O\left(\frac{1}{n^3}\right)$$ making $$a_n\simeq \frac 1{\sqrt{\pi n}}e^{-\frac 1 {8n}} < \frac 1{\sqrt{\pi n}}$$