MHB Is it possible to prove the inequality without using induction?

[anemone]

Show that $\dfrac{1}{2} \cdot \dfrac{3}{4} \cdot \dfrac{5}{6} \cdots \dfrac{1997}{1998} >\dfrac{1}{1999}$, where the use of induction method is not allowed.

 

[Ackbach]

Spoiler

This is equivalent to showing
$$1 \cdot 3 \cdot 5 \cdot \dots \cdot 1999 > 2 \cdot 4 \cdot 6 \cdot \dots \cdot 1998.$$
But this is so, since $3>2$, and $5>4$, etc. The left-hand product has one more term (the number $1$) in it than the right-hand term. Therefore, the left-hand product is greater.

 

[anemone]

Wow...the moment I saw this problem, I naturally perceived as a hard to prove sort of inequality problem but you cracked it in such an elegant method! Well done, Ackbach, and thanks for participating! :)