For all integers $n\geq 1$, $$\frac{1}{2}\cdot \frac{3}{4} \cdot\cdots\cdot \frac{2n-1}{2n} \leq \frac{1}{\sqrt{3n + 1}}.$$
Need help with inductive step.
IB: P(1) $$\prod_{i=1}^{1} \frac{2i-1}{2i} = \frac{1}{2} \leq \frac{1}{\sqrt{3(1)+1}} = \frac{1}{\sqrt{4}}$$ IH: For some $k \geq 1$, assume P(K) is true. P(K) = $$\prod_{i=1}^{k} \frac{2i-1}{2i} \leq \frac{1}{\sqrt{3k+1}}$$
the beginning: $n=1$ then we have $\frac{1}{2}\le \frac{1}{\sqrt{4}}$ and this is true. assuming for $n=k$ we have $$\frac{1}{2}\cdot \frac{3}{4}\cdot ...\cdot \frac{2k-1}{2k}\le \frac{1}{\sqrt{3k+1}}$$ and we have to prove that for $n=k+1$ is hold: $$\frac{1}{2}\cdot \frac{3}{4}\cdot ...\cdot \frac{2k-1}{2k}\cdot \frac{2k+1}{2(k+1)}\le \frac{1}{\sqrt{3k+4}}$$ we have to Show that $$\frac{1}{\sqrt{3k+1}}\cdot \frac{2k+1}{2(k+1)}\le \frac{1}{\sqrt{3k+4}}$$ which is true. the last inequality is given by squaring as $$(3k+4)(2k+1)^2\le 4(k+1)^2(3k+1)$$
For the inductive step, note that that for $n\geq 1$, $$\prod_{i=1}^{n+1} \frac{2i-1}{2i}= \prod_{i=1}^{n+1} \frac{2i-1}{2i} \cdot \frac{2(n+1)-1}{2(n+1)}\leq \frac{1}{\sqrt{3n + 1}}\cdot \frac{2n+1}{2n+2}.$$ So it remains to show that $$\frac{1}{\sqrt{3n + 1}}\cdot \frac{2n+1}{2n+2}\leq \frac{1}{\sqrt{3(n+1) + 1}}.$$ that is $$(3n+4)(2n+1)^2 \leq (3n+1)(2n+2)^2.$$ Can you take it from here?
