I am self studying analytic number theory from class notes of a senior and in it I am unable to deduce an inequality which is not proved .
Assume $b(m) = \sum_{s, t} \frac{1} {log(s) log(t) } $with these 3 conditions:
- $2\leq s \leq n$
- $2\leq t \leq n$
- $s+t= m $
It is to be proved that if $m \leq n$, then $b(m) \geq (m-3) log^{-2} n $
Please help.
If $n\ge m$ the sum consists of $m-3$ terms. Since $\log x$ for $x\ge2$ is positive increasing function $$ \frac1{\log s \log (m-s)}>\frac1{\log^2 m}.$$
Therefore: $$b(m)>\frac{m-3}{\log^2m}\ge\frac{m-3}{\log^2n}.$$