 Apsidal speed sum for an attractive inverse-square orbit

For unit mass with inverse-square potential $-Q/r$, $Q>0$, and positive conserved angular momentum $h$, a bounded orbit has $-Q^2/(2h^2)\le E<0$. Its periapsis distance and apoapsis distance are roots of $|E|r^2-Qr+h^2/2=0$, so their sum is $Q/|E|$ and product $h^2/(2|E|)$. Speed is largest at the periapsis and smallest at the apoapsis, because $v^2=2(E+Q/r)$ decreases with $r$. Radial velocity is zero there, so the two speeds are $h/r_{\min }$ and $h/r_{\max }$. Their sum is $h(r_{\min }+r_{\max })/(r_{\min }{r}_{\max })=2Q/h$, independent of $E$, including the circular limit.