Inequality Challenge: Prove 1/44 > 1/1999

[anemone]

Show that $$\frac{1}{44}>\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\left(\frac{5}{6}\right)\cdots\left( \frac{1997}{1998}\right)>\frac{1}{1999}$$

 

[mathworker]

For the first one,
$$\left(\frac{1+1}{2+1}\right)\cdots\Big(\frac{1997+1}{1998+1}\Big)>\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)\Big(\frac{5}{6}\Big)\cdots\Big(\frac{1997}{1998}\Big)
$$
$$\Big(\frac{2}{3}\Big)\Big(\frac{4}{5}\Big)\cdots \Big(\frac{1998}{1999}\Big)>\Big(\frac{1}{2}\Big).\Big(\frac{3}{4}\Big)\Big(\frac{5}{6}\Big)\cdots\Big(\frac{1997}{1998}\Big)$$
$\text{let,}$
$$\frac{1}{2}\frac{3}{4}\frac{5}{6}\cdots\frac{1997}{1998}=x$$
$$\frac{1}{x\cdot1999}>x$$
$$x^2<\frac{1}{1999}$$
$$x<\frac{1}{44+x}$$
$\text{hence,}$
$$x<\frac{1}{44}$$

 

[Opalg]

Show that $$\frac{1}{44}>\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\left(\frac{5}{6}\right)\cdots\left( \frac{1997}{1998}\right)>\frac{1}{1999}$$

The other inequality is easily proved by induction. Let $P_n = \Bigl(\dfrac{1}{2}\Bigr)\Bigl(\dfrac{3}{4}\Bigr) \cdots \Bigl(\dfrac{2n-1}{2n}\Bigr).$ Then $P_n > \dfrac{1}{2n+1}$.

Proof. The base case $n=1$ certainly holds. Assuming the inductive hypothesis, $$P_{n+1} = P_n\Bigl(\dfrac{2n+1}{2n+2}\Bigr) > \dfrac{1}{2n+1}\Bigl(\dfrac{2n+1}{2n+2}\Bigr) = \dfrac{1}{2n+2} > \dfrac{1}{2n+3},$$ which completes the inductive step.