Let $n$ be a positive integer and $p$ the largest prime $\le n$. Let $L$ denote a subset of the positive composite numbers less than $n$. Consider all integers $i$ with no prime factors larger than $p$: $$i=2^{a_2}3^{a_3}\cdots p^{a_p}; a_2,a_3,\ldots,a_p\in \mathbb{Z}_{\ge 0}.$$ Note that $i$ may be larger than $n$. Is it true that for each $m$ and each $L$ of size $m$, $$\bigg(\prod_{p\le n,p\,\text{prime}}\frac{p-1}{p}\bigg)\sum_{i: \frac{j}{\gcd(i,j)}\text {composite}\, \forall j\in L}\frac{1}{i}\le \frac{n-m}{n}\,\,?$$
I can show this inequality for small values of $n$, but in the general case I am having difficulty. The main difficulty I've had is describing the set of $i$'s for which $\frac{j}{\gcd(i,j)}$ is composite for all $j\in L$ ( Finding all integers $i$ such that $\frac{j}{\gcd(i,j)}$ is composite for some $j's$.). Perhaps there is an upper bound that can be placed on the sum that still gives the inequality.
Example: Consider $n=6$. Then $m=1$ or $2$ since $4,6$ are the only composites $\le 6.$ If $m=2$, then $L=\lbrace 4,6\rbrace.$ In order for $\frac{j}{\gcd(i,j)}$ to be composite for $j=4,6$, we must have that $i$ is not divisible by 2 nor 3, so $i=5^{a_5}.$ The left hand side is $\frac{2-1}{2}\frac{3-1}{3}\frac{5-1}{5}\sum_{a_5\ge 0}\frac{1}{5^{a_5}}=\frac{1}{2}\frac{2}{3}=1/3$, while the right hand side is $\frac{6-2}{6}=2/3.$
Let $$F_k(s) = \prod_{p \le k} \frac{1}{1-p^{-s}}= \sum_{n=1}^\infty n^{-s} 1_{\text{Lpf}(n) \le k}, \qquad G_k(s) = \prod_{p > k} \frac{1}{1-p^{-s}}=\sum_{n=1}^\infty n^{-s} 1_{\text{lpf}(n) > k}$$
$\qquad$ where $\text{Lpf}(n)= \sup_{p | n} p,\quad \text{lpf}(n)= \inf_{p | n} p$.
And $$H_k(s) = F_k(s)(G_k(s)-1-\sum_{p > k} p^{-s}) = \sum_{n=1}^\infty h_k(n) n^{-s}$$ Can you describe the coefficients $h_k(n)$ ?
Can you evaluate $H_k(s)$ and $G_k(s)-1-\sum_{p > k} p^{-s}$ at $s=1$ ?