# Need help in proving an inequality related to logarithms

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:

1. $$2\leq s \leq n$$
2. $$2\leq t \leq n$$
3. $$s+t= m$$

It is to be proved that if $$m \leq n$$, then $$b(m) \geq (m-3) log^{-2} n$$

Please help.

• If $m\le n$, then $n$ should not play any role. – user Apr 26 '20 at 20:21
• @user can you please write an answer which is a bit more ellaborated . I am confused. – Tim Apr 26 '20 at 20:35
• It is not an answer but a comment aimed on improvement of your question. Do you see that if $n\ge m$, it can be replaced with $m-2$? – user Apr 26 '20 at 20:42
• @user m -2 is not greater than Or equal to m. So, how can you replace it? – Tim Apr 26 '20 at 20:46
• @user also why do you want to " improve the question " ? Do you mean giving a hint for proof? – Tim Apr 26 '20 at 20:47

## 1 Answer

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: $$b(m)>\frac{m-3}{\log^2m}\ge\frac{m-3}{\log^2n}.$$