In some paper I saw that to show that some inequality of this type $$ \sum_{p\leq X}\frac{f(p)}p>\frac{\log^2X}{C_1} $$ it says the above inequality followes using by partial summation if we show $$ \sum_{p\leq X}f(p)\log p>\frac{2X\log^2X}{C_2}. $$ My question is "is it a general method in analytic number theory or it depends on the problem and the estimates that we can do easier in the second inequality in that problem"?
Thanks.
It is a consequence of the Abel's summation. We have that $$\sum_{p\leq x}\frac{f\left(p\right)}{p}=\sum_{p\leq x}f\left(p\right)\log\left(p\right)\frac{1}{p\log\left(p\right)} $$ $$=\frac{1}{x\log\left(x\right)}\sum_{p\leq x}f\left(p\right)\log\left(p\right)+\int_{2}^{x}\frac{\sum_{p\leq t}f\left(p\right)\log\left(p\right)\left(\log\left(t\right)+1\right)}{t^{2}\log^{2}\left(t\right)}dt $$ so if $$\sum_{p\leq x}f\left(p\right)\log\left(p\right)>\frac{2x\log^{2}\left(x\right)}{C_{2}} $$ we have $$\sum_{p\leq x}\frac{f\left(p\right)}{p}\geq\frac{2\log\left(x\right)}{C_{2}}+\frac{2}{C_{2}}\int_{2}^{x}\frac{\log\left(t\right)+1}{t}dt $$ $$=\frac{4\log\left(x\right)}{C_{2}}+\frac{\log^{2}\left(x\right)}{C_{2}}+C_{3}>\color{red}{\frac{\log^{2}\left(x\right)}{C_{2}}}$$