<s>[INST] I have troubles with proving the induction step: $\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}}$ Let's assume that the unequality is right if $n = k$, that is:

$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k-1)}{2k}  ≤  \frac{1}{\sqrt{2k+1}}$

Let's prove that the unequality is right for $n = k + 1$ as well, that is:

$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2}  ≤  \frac{1}{\sqrt{2k+3}}$

I've spent a comparable amount of time trying to prove this point but looks like I can't grasp what to do.
 [/INST] Note that$$\frac12\times\frac34\times\frac56\times\cdots\times\frac{2k-1}{2k}\times\frac{2k+1}{2k+2}\leqslant\frac1{\sqrt{2k+1}}\times\frac{2k+1}{2k+2},$$by the induction hypothesis. So, it would be great if you could prove that$$\frac1{\sqrt{2k+1}}\times\frac{2k+1}{2k+2}\leqslant\frac1{\sqrt{2k+3}}\tag1$$But $(1)$ is equivalent to$$\left(\frac{2k+1}{2k+2}\right)^2\leqslant\frac{2k+1}{2k+3},$$which in turn is equivalent to$$(2k+1)(2k+3)\leqslant(2k+2)^2.$$Can you take it from here?
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