<s>[INST] To prove $\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots \frac{(2n-1)}{2n} \le \frac{1}{\sqrt{3n+1}}$ To prove $$P=\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots  \frac{(2n-1)}{2n}\le \frac{1}{\sqrt{3n+1}}$$
i have written $P$ as
$$P=\frac{(2n)!}{2^{2n}(n!)^2}=\frac{(2n)!}{4^{n}(n!)^2}=\frac{\binom{2n}{n}}{4^n}$$
Now
$$P=\frac{\binom{2n}{n}}{(1+3)^n} \lt \frac{\binom{2n}{n}}{1+3n}$$  since
$$(1+3)^n=1+3n+\binom{n}{2}3^2+\cdots$$
Any help here..
 [/INST] We can use simple induction to find the answer. First, at $n=1$, the inequality holds true. Now let us assume the above for n and prove it for $n+1$.
Let us divide the LHS of the inequality for $n+1$ by LHS of the inequality for n, and do the same with the RHS for $n+1$ and $n$.
$(2n+1)/(2n+2)$ must be lesser than or equal to $\sqrt{(3n+1)}/\sqrt{(3n+3)}$. We can square on both sides.
$(4n^2+4n+1)/(4n^2+8n+4)$ must be lesser than or equal to $(3n+1)/(3n+3)$.
$(4n^2+4n+1)(3n+3)$ must be lesser than or equal to $(4n^2+8n+4)(3n+1)$. This can be easily verified by multiplying the polynomials. Hence, for all '$n$', the inequality holds
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