The title basically says it all, we've got to prove the inequality without calculating the value of the left side: $$\prod_{n = 2}^{40}\log_{2n - 1}(2n) > 2$$ aka $$\log_34\cdot\log_56\cdot\log_78\cdot\ldots\cdot\log_{79}80 > 2$$
Everything I've tried eventually brings be back to where I started... I'm obviously missing the key here and for god's sake I can't think of anything new anymore! :D
Every suggestion is welcome. Thank you in advance!
Note that $$ \log_34>\log_45\\ \log_56>\log_67\\ \vdots $$ So $$ (\log_34\cdot\log_56\cdots\log_{79}80)^2>\log_34\cdot\log_45\cdots\log_{79}80\cdot\log_{80}81 $$ The right-hand side is equal to $4$. This is easiest to see if you raise $3$ to the power of the right-hand side and use the logarithms one by one until you end up with $81=3^4$.