Prove by induction a product inequality...

Question

Prove by induction that for every natural number is verified that:

$$ \prod_{i=1}^n \frac{2i}{2i + 3} < \frac{2}{(n+1)\sqrt{2n+4}} $$ I get in trouble when applying the hypothesis to proof for $n+1$, it looks like I'm getting to nowhere.

Thank you.

Answer 1

Inductive step: $$\prod_{i=1}^{n+1} \frac{2i}{2i + 3}=\prod_{i=1}^n \frac{2i}{2i + 3}\cdot \frac{2(n+1)}{2(n+1)+3}<\\ <\frac{2}{(n+1)\sqrt{2n+4}}\cdot \frac{2(n+1)}{2(n+1) + 3} < \frac{2}{((n+1)+1)\sqrt{2(n+1)+4}} \iff \\ \frac{2}{\sqrt{n+2}}\cdot \frac{1}{2n+5} < \frac{1}{(n+2)\sqrt{n+3}} \iff \\ \frac{2}{2n+5} < \frac{1}{\sqrt{(n+2)(n+3)}} \iff \\ 4(n+2)(n+3)<(2n+5)^2 \iff \\ 0<1.$$

Answer 2

In explicit terms, the LHS is given by $\frac{3\cdot 4^n}{(2n+3)(2n+1)\binom{2n}{n}}$, hence the given inequality is equivalent to

$$ \frac{(2n+3)(2n+1)\binom{2n}{n}}{3\cdot 4^n}>\frac{(n+1)\sqrt{n+2}}{\sqrt{2}} $$ or to

$$ \left[\frac{1}{4^n}\binom{2n}{n}\right]^2>\frac{9(n+1)^2(n+2)}{2(2n+3)^2(2n+1)^2}. $$ The inequality holds as an equality at $n=0$ and the ratio between the LHS and the RHS is an increasing function over $\mathbb{N}$. Indeed by letting $$ f(n) = \frac{2(2n+3)^2(2n+1)^2}{9(n+1)^2(n+2)}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2 $$ we have $$ \frac{f(n+1)}{f(n)}=\frac{(2n+5)^2}{(2n+4)(2n+6)}>1. $$