Attempt to prove an inequality by induction I am trying to prove an inequality by induction which is as follows:
$$\frac{(2n)!}{2^{2n}\cdot (n!)^2} \le \frac{1}{\sqrt{3n + 1}}$$
Base Case, i.e, for n =1,
$$\frac{(2(1))!}{2^{2(1)}\cdot ((1)!)^2} \le \frac{1}{\sqrt{3(1) + 1}}$$
$$\implies \frac{2!}{2^{2}\cdot (1)^2} \le \frac{1}{\sqrt{3 + 1}}$$
$$\implies \frac{2}{4} \le \frac{1}{2}$$
$$\implies \frac{1}{2} \le \frac{1}{2}$$
which is true
Assume that the inequality is true for $n=m$ where $m \in \mathbb{N}$   , i.e,
$$\frac{(2m)!}{2^{2m}\cdot (m!)^2} \le \frac{1}{\sqrt{3m + 1}}$$
$$\implies \quad \quad \quad\quad\quad\quad\quad \frac{1\cdot2\cdot3\cdot .....\cdot (2m-1)\cdot (2m)}{2^{2n}\cdot (1\cdot2\cdot3\cdot .....\cdot (m-1) \cdot m)^2} \le \frac{1}{\sqrt{3m + 1}} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$
$$\implies \quad \quad \quad  \frac{1\cdot2\cdot3\cdot .....\cdot (2m-1)\cdot (2m) \cdot(2m+1)\cdot(2m +2)}{2^{2m}\cdot (1\cdot2\cdot3\cdot .....\cdot (m-1) \cdot m)^2\cdot(2m+2)\cdot (m+1)\cdot2} \le \frac{1}{\sqrt{3m+ 1}}\cdot \frac{(2m +1) }{(m+1)\cdot 2} \quad\quad\quad\quad\quad\quad\quad\quad$$
$$\implies \frac{(2m+2)!}{2^{2m +2}\cdot (1\cdot2\cdot3\cdot .....\cdot (m-1) \cdot m \cdot (m+1))^2 }\le \frac{1}{\sqrt{3m+ 1}}\cdot \frac{(2m +1) }{(m+1)\cdot 2}$$
$$\implies \frac{(2(m+1))!}{2^{2(m +1)}\cdot ((m+1)!)^2 }\le \frac{1}{\sqrt{3m+ 1}}\cdot \frac{(2m +1) }{(m+1)\cdot 2}$$
$$\implies \frac{(2(m+1))!}{2^{2(m +1)}\cdot ((m+1)!)^2 }\le \frac{1}{\sqrt{12m+ 4}}\cdot \frac{(2m +1) }{(m+1)}\quad \quad ......(i)$$$$
Now , I want $LHS$ to be smaller than or equalto $\frac{1}{\sqrt{3m+4}}$. For this I thought to prove that
$$\frac{1}{\sqrt{12m+ 4}}\cdot \frac{(2m +1) }{(m+1)} \le \frac{1}{\sqrt{3m+4}} \quad \quad ......(ii)$$
Now ,
$$12m>3m$$
$$12m + 4 > 3m +4$$
$$\sqrt{12m +4} > \sqrt{3m +4}$$
$$\frac{1}{\sqrt{12m +4}} < \frac{1}{\sqrt{3m +4}}$$
Then, I thought that if $\frac{(2m +1) }{(m+1)}$ is smaller than $1$ , i would get inequality (i) but this is false. So I am lost.
Can anyone help me to proceed further to get the desired result?
Thanking in advance

I think you messed something up in the computation.
If you want to prove this using induction, what you need to show is that
\begin{equation*}
\frac{1}{\sqrt{3n+1}}\frac{(2n+1)(2n+2)}{4(n+1)^2}\leq \frac{1}{\sqrt{3n+4}}.
\end{equation*}
After simplifying this means
\begin{equation*}
\frac{(2n+1)^2}{(2n+2)^2}\leq \frac{3n+1}{3n+4},
\end{equation*}
which you can easily check by direct computation.
