Inequality $\binom{2n}{n}\leq 4^n$

A bit more can be proven with a bit more work. For $k\ge0$, we have the inequality $$ \begin{align} \left(\frac{k+\frac12}{k+1}\right)^2 &=\frac{k^2+k+\frac14}{k^2+2k+1}\\ &\le\frac{k+1}{k+2}\tag{1} \end{align} $$ because cross-multiplication gives $k^3+3k^2+\frac94k+\frac12\le k^3+3k^2+3k+1$.

Using $(1)$ yields $$ \begin{align} \frac{\binom{2k+2}{k+1}}{\binom{2k}{k}} &=4\frac{k+\frac12}{k+1}\\ &\le4\sqrt{\frac{k+1}{k+2}}\tag{2} \end{align} $$ Multiplying $(2)$ for $k=0$ to $k=n-1$, we get $$ \boxed{\bbox[5px]{\displaystyle\binom{2n}{n}\le\frac{4^n}{\sqrt{n+1}}}}\tag{3} $$

As Olivier Oloa comments, Stirling's Formula tells us that $$ \lim_{n\to\infty}\binom{2n}{n}\frac{\sqrt{\pi n}}{4^n}=1\tag{4} $$ In fact, using inequalities similar to $(2)$, in this answer, it is shown that $$ \boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi(n+\frac13)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}}}}\tag{5}$$

$\displaystyle 4^n=(1+1)^{2n} =\sum_{k=0}^{2n}\binom{2n}{k}\geq\binom{2n}{n}$

Let $n \in \mathbb{N}.$ You may write

$$ 4^n=2^{2n} =(1+1)^{2n}= \sum_{k=0}^{2n}\binom{2n}{k}\geq\binom{2n}{n}. $$