Note on translated sum on primitive sequences

In this note, we construct a new set \boldsymbol{S} of primitive sets such that for any real number x\geq 60 get: \begin{equation*} \sum\limits_{a\in \mathcal{A}}\frac{1}{a(\log a+x)}>\sum\limits_{p\in \mathcal{P}}\frac{1}{p(\log p+x)},\text{ }\mathcal{A\in }{\boldsymbol{S}}, \end{equation*} where \mathcal{P} denotes the prime numbers.