Show that this inequality is true Show that $\frac{2}{3} \cdot \frac{5}{6} \cdot \frac{8}{9} \cdot ... \cdot \frac{999998}{999999} > \frac{1}{100}$.
I tried to take another multiplication $\frac{3}{5} \cdot \frac{6}{8} \cdot \frac{9}{11} \cdot ... \cdot \frac{999996}{999998}$ so that we would have their multiplications equal to $\frac{2}{999999}$. And if we assume that first one equals to $x$, second one equals $y$ we would have an inequality like $x \gt y$ and $x^2 \gt y \cdot x$ so that we can prove that $ x \gt \frac{1}{1000}$ but I can't make it for $\frac{1}{100}$.

Let $$A=\frac{2}{3}\cdot\frac{5}{6}\cdot...\cdot\frac{999998}{999999},$$
$$B=\frac{1}{2}\cdot\frac{4}{5}\cdot...\cdot\frac{999997}{999998}$$ and $$C=\frac{3}{4}\frac{6}{7}\cdot...\cdot\frac{999999}{1000000}.$$
Thus, since $$\left(\frac{3n+2}{3n+3}\right)^2>\frac{3n+1}{3n+2}\cdot\frac{3n+3}{3n+4},$$ we obtain: $$A^2>BC,$$ which gives
$$A^3>ABC=\frac{1}{1000000}.$$
