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For $\psi(d)=d\prod_{p\mid d}(1+1/p)$ and every $q>0$,

$$\sum_{\psi(d)/d=q}\frac1d\le1. \tag{1}$$

Every nonempty fibre has a unique finite prime support, and its mass is $\prod_{p\in P}(p-1)^{-1}$. Equality in (1) occurs at $q=1$ and $q=3/2$, corresponding to $P=\varnothing$ and $P=\{2\}$.

Proof. For any finite support $P$,

$$q=\prod_{p\in P}\frac{p+1}{p}.$$

If its largest prime $r$ is at least five, $r$ cannot divide any numerator factor $p+1$ with $p\le r$. The only possible equality $p+1=r$ would make $p=r-1$ an even integer greater than two. Also every prime factor of $p+1$ is less than $r$: for odd $p$, it is at most $(p+1)/2<r$, and the factor from $p=2$ is three. Thus $r$ is exactly the largest prime of the reduced denominator. Remove its factor $(r+1)/r$ and continue with a smaller support.

If the reduced denominator has no prime at least five, no support prime at least five can remain. The four possible supports and their ratios are exactly

$$\begin{array}{c|c|c} P&q&\prod_{p\in P}(p-1)^{-1}\\ \hline \varnothing&1&1\\ \{2\}&3/2&1\\ \{3\}&4/3&1/2\\ \{2,3\}&2&1/2 \end{array}$$

Their ratios are distinct. This supplies a unique base case for the descent, including the reduced denominator one at $q=2$. Consequently any ratio having a solution determines exactly one support. All positive exponent choices on that support give the same ratio, and no other integer does. Summing their reciprocals as in Lemma 2.1 gives the claimed product. Each factor is at most one; equality forces the support to be empty or just $\{2\}$. Empty fibers contribute zero. $\square$

This supplies the finite case analysis and the induction left to the reader in source Remark 4.7. In particular, the cancellation at $\{2,3\}$ is retained rather than silently using the totient base case.

Source. Tao, published paper, published pp.818–819, Remark 4.7. This page uses that published version.

Bears on. Problem 49.