<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] Using the induction hypothesis:
$$\frac{2k+1}{2k+2} ≤ \frac{\sqrt{2k+1}}{\sqrt{2k+3}}$$
Now let $u = 2k+2$:
$$\frac{u-1}{u} ≤ \frac{\sqrt{u-1}}{\sqrt{u+1}}$$
For $u ≥ 1$ or $2k + 2 ≥ 1$ which implies $k ≥ -\frac{1}{2}$, both sides are positive. So we can square both sides:
$$\frac{u^2-2u+1}{u^2} ≤ \frac{u-1}{u+1}$$
$$u^3-u^2-u+1 ≤ u^3-u^2$$
$$u ≥ 1$$
Obviously this is not a formal proof. To do so, you need to invert all the steps starting from $u ≥ 1$.
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