$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$. Show that
$$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})<\frac{1}{10}$$
My attempt: This  problem is from a text book where is introduced as: https://en.wikipedia.org/wiki/Generalized_mean 
This wouldn't be a problem if I knew the sum or a product of the given numbers. Then I will use AG inequality, but I don't know how. 

Let $a=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}. . . .\frac{99}{100}$
and:
$b=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}. . . \frac{100}{101}$
It is clear that $a<b$ because each factor of $a$ is less than its corresponding factor in $b$ :
$\frac{1}{2}<\frac{2}{3}, \frac{3}{4}<\frac{4}{5}. . . \frac{99}{100}<\frac{100}{101}$
⇒ $a^2 < ab=(\frac{1}{2}.\frac{2}{3}).(\frac{3}{4}.\frac{4}{5}). . . .(\frac{99}{100}.\frac{100}{101})=\frac{1}{101}$
$a^2<\frac{1}{101}$⇒$a<\frac{1}{\sqrt{101}}<\frac{1}{10}$
Also:
$2a=\frac{3}{4}.\frac{5}{6}. . . .\frac{99}{100}$
$\frac{3}{2}.b=\frac{4}{5}.\frac{6}{7}. . . \frac{100}{101}$
$2a<\frac{3}{2}b$⇒$2a^2<\frac{3}{2}ab=\frac{3}{2}.\frac{3}{101}$ 
Or $a^2<\frac{3}{4}ab=\frac{3}{4}.\frac{3}{101}$
Since $9>4$ then  $a^2 >\frac{4}{9\times 101}$ and therefore:
$a>\frac{1}{15}$
A more reliable reasoning is given as a comment for this part:
$2a>b$ ⇒ $2a^2>ab=\frac{1}{101}$⇒$a^2>\frac{1}{202}$⇒$a>\frac{1}{\sqrt{202}}>\frac{1}{\sqrt{225}}=\frac{1}{15}$
