Let $D_b(m)$ be the number of divisors of $m$ that are less than $b$. Neil Sloane has suggested that the number of binary quadratic forms $Ax^2 + Bxy + Cy^2$ with integral coefficients, discriminant $n^2-4$, and satisfying $A > 0$, $C > 0$, and $B > A+C$ is equal to
$$ \sum_{b=1}^{n-1} D_b(b^2-nb+1) $$
For a given integer $b \geq 1$, let $d_b(m)$ be the number of ways to nontrivially factor $m$ into two factors that are both congruent to 1 modulo $b$. It turns out that for each integer $n \geq 1$, the sum
$$ \sum_{b=1}^{n-1} (d_b(b^2+nb+1) - 2) $$
is equal to the number of binary quadratic forms of discriminant $n^2-4$ of the form $(st+1)x^2 + (rst+r+s+t)xy + (rs+1)y^2$ with integers $r$, $s$, $t > 0$. Since these forms are a subset of the earlier set of forms, we have the inequality
$$ \sum_{b=1}^{n-1} D_b(b^2-nb+1) > \sum_{b=1}^{n-1} (d_b(b^2+nb+1) - 2) $$
The visual similarity of both sides is striking. Is there a different way, especially a more direct way, to prove this inequality?
(I note that the individual numbers $D_b (b^2-nb+1)$ are sometimes lower and sometimes greater than the individual numbers $d_b (b^2+nb+1) - 2$.)