= Solution
Write the <gravitational potential> with <spherical symmetry> as $\Phi(r)$ and let $\mathbf n$ be the unit <vector> normal to a <circular orbit>, oriented in the direction of its <angular velocity>. Put $\boldsymbol\Omega=\Omega\mathbf n$ and denote the <specific angular momentum> by $\mathbf h=h\mathbf n$. Circular force balance and the <specific orbital energy> give
$$
\Phi'(r)=r\Omega^2,\qquad h=r^2\Omega,\qquad \varepsilon=\Phi(r)+\frac12r^2\Omega^2.
$$
Differentiating along the family of <circular orbits>,
$$
\frac{d\varepsilon}{dr}=2r\Omega^2+r^2\Omega\Omega'=\Omega\frac{dh}{dr}.
$$
A change in orbital orientation contributes $h\,d\mathbf n$ to $d\mathbf h$, but $\mathbf n\cdot d\mathbf n=0$. Consequently
$$
\boxed{d\varepsilon=\Omega\,dh=\boldsymbol\Omega\cdot d\mathbf h.}
$$
This <circular-orbit energy gradient> accounts for radial and inclination changes together; <spherical symmetry> makes the orbital <energy> independent of the plane.
For either variable-mass particle, $\mathbf H=M\mathbf h$ and $E=M\varepsilon$. Thus $d\mathbf H=M\,d\mathbf h+\mathbf h\,dM$, so
$$
dE=\varepsilon\,dM+M\boldsymbol\Omega\cdot d\mathbf h=(\varepsilon-\Omega h)dM+\boldsymbol\Omega\cdot d\mathbf H.
$$
<Conservation of mass> and <conservation of angular momentum> impose $dM_2=-dM_1$, $d\mathbf H_2=-d\mathbf H_1$. Adding the two particle contributions proves
$$
\boxed{dE=\{(\varepsilon_1-\Omega_1h_1)-(\varepsilon_2-\Omega_2h_2)\}\,dM_1+(\boldsymbol\Omega_1-\boldsymbol\Omega_2)\cdot d\mathbf H_1.}
$$
To interpret the signs, label the inner orbit 1 and the outer orbit 2. For $f(r)=\varepsilon-\Omega h$ the first identity gives $f'=-h\Omega'$. The assumed outward decrease of <angular velocity> therefore makes $f_1<f_2$, so the mass-exchange part lowers the <energy> when $dM_1>0$: <mass> is transferred inward. In a coplanar, co-rotating pair, $\Omega_1>\Omega_2$ and an outward <angular momentum> transfer has $d\mathbf H_1=-dH\mathbf n$ with $dH>0$. Its contribution is $-(\Omega_1-\Omega_2)dH<0$.
The <vector> difference also identifies the inclination-reducing exchange. For a fixed small magnitude of admissible <angular momentum> exchange, its <energy> contribution is most negative when
$$
d\mathbf H_1=-s(\boldsymbol\Omega_1-\boldsymbol\Omega_2),\qquad s>0,
$$
by the <Cauchy-Schwarz inequality>. This gives $dE_{\rm angular}=-s|\boldsymbol\Omega_1-\boldsymbol\Omega_2|^2$. Let $q=\mathbf n_1\cdot\mathbf n_2=\cos\iota$ and $H_i=|\mathbf H_i|$, taking fixed particle masses for this exchange. Projecting $d\mathbf H_i$ perpendicular to its own direction gives
$$
d\mathbf n_1=\frac{s\Omega_2}{H_1}(\mathbf n_2-q\mathbf n_1),\qquad d\mathbf n_2=\frac{s\Omega_1}{H_2}(\mathbf n_1-q\mathbf n_2).
$$
Therefore
$$
dq=s(1-q^2)\left(\frac{\Omega_2}{H_1}+\frac{\Omega_1}{H_2}\right)>0\qquad(0<\iota<\pi).
$$
The downhill exchange reduces the relative <orbital inclination>. Its inner <angular momentum> component is $\mathbf n_1\cdot d\mathbf H_1=-s(\Omega_1-\Omega_2q)<0$; in the aligned limit the outer orbit receives that <angular momentum> directly. For strongly inclined orbits, part of the exchange instead cancels differently directed angular momenta as the planes align. Together these gradients explain the tendencies toward inward <mass> transfer, outward <angular momentum> transport and coplanarity.
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