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.
On
Note that the induction hypotesis reduces to show that
$$ \frac{1}{\sqrt{2k+1}} \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}} \implies\sqrt{(2k+1)(2k+3)}\le 2k+2$$
Can you finish from here?
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Note that if $$\frac{a}{b}<1 \Rightarrow \frac{a}{b} < \frac{a+x}{b+x}, x>0.~~~(1)$$ Let $$P_1=\frac{1}{2}~ \frac{3}{4}~ \frac{5}{6}....\frac{2k+1}{2k+2}.$$ Let us add 1, up and down in each fraction to get $$P_2=\frac{2}{3} ~\frac{4}{5} ~\frac{6}{7}...\frac{2k+2}{2k+3}.$$ So By (1), $$P_1<P_2 \Rightarrow P_1^2 < P_1. P_2 \Rightarrow P_1^2 < \frac{1}{2k+3}$$$$\Rightarrow P_1 <\frac{1}{\sqrt{2k+3}}.$$
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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 $u ≥ 1$.
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?