Show that this inequality is true

Question

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 $\frac{1}{100}$.

Answer 1

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 $$A^3>ABC=\frac{1}{1000000}.$$

Answer 2

Let $$A_n=\sqrt[3]{3n+1}\cdot\prod_{k=1}^n\left(1-\frac1{3k}\right).$$ The claim is that $$\tag1 A_{333333}>1.$$ We compute $$\begin{align}\left(\frac{A_{n}}{A_{n-1}}\right)^3&=\frac{(3n+1)(3n-1)^3}{(3n-2)27n^3}\\ &=1+\frac{6n-1}{(3n-2)27n^3}\\&>1+\frac{6n-4}{(3n-2)27n^3}\\&=1+\frac2{27n^3}>1\end{align}$$ or $A_n>A_{n-1}$. By induction, $$A_{333333}>A_1=\sqrt[3]{4}\cdot \frac23 =\sqrt[3]{\frac{32}{27}}>1$$