added, Sept 20, 2010. The bounty is for an answer to either question 1 or 2.

I've made partial progress toward 2 using variational methods(direct method of the calculus of variations). I can prove that if a syzygyis chosen anywhere in a neighborhood of binary collision (so $r_{12}(t_c) = \delta$, small, $r_{23} (t_) = r_{13}(t_c) + \delta$)then there exists a brake orbit solutionarc ending in this syzygy and satisfying the inequality of question 2. The proof suggests, but does not prove, that the result holds locally nearisosceles, meaning for brake initial conditionsin a neighborhood of isosceles brake initial conditions ( so
$r_{13} (0) = r_{12} (0) + \epsilon$). If I had uniqueness [modulo rotation and reflection] of brake orbits with specified syzygy endpoints, then my proof would yield a proof of this local version of the alleged theorem.
Unfortunately, my proof does notexclude the possibility of more than one orbit ending in the chosen syzygy, one of which violates the inequality.

Consider the usual three-body problem with Newtonian $1/r^2$ force between masses. Let the three masses start off at rest, and not collinear. Then they will become collinear a finite time later by a theorem I proved some time ago. (See the papers Infinitely Many Syzygies' andThe zero angular momentum three-body problem: all but one solution has syzygies' available on my web site or the arXivs.) Let $t_c$ denote the first such time.

Write $r_{ij} (t)$ for the distance between mass $i$ and mass $j$ at time $t$.

Question 1. For general masses $m_i >0$, is it true that the `moment of inertia' $I = m_1 m_2 r_{12} ^2 + m_2 m_3 r_{23} ^2 + m_1 m_3 r_{13} ^2$ monotonically decreases over the interval $(0, t_c)$?

Question 2. If the masses are all equal and if the initial side-lengths satsify $0 < r_{12}(0) < r_{23} (0)< r_{13} (0)$
is it true that these inequalities remain in force: $0 < r_{12} (t) < r_{23} (t) < r_{13}(t)$ for $0 < t < t_c$? In other words: if the triangle starts off as scalene (not isosceles, and of having nonzero area) does it remain scalene up to collinearity?

Question 2. For general masses, is it true that the `moment of inertia' $I = m_1 m_2 r_{12} ^2 + m_2 m_3 r_{23} ^2 + m_1 m_3 r_{13} ^2$ monotonically decreases over the interval $(0, t_c)$?

Motivation: The space of collinear triangles, consisting of triangles of zero area, acts like a global Poincare section for the zero-angular momentum, negative energy three-body problem. To understand obtain some understanding of the "return map " from this "syzygy plane" space to itself the "brake orbits"-``brake orbits''-- those solutions for which all velocities vanish at some instant -- seems seem to play an organizing role. Answering either questions would yield useful information about brake orbits.

Aside: I suspect that if the answers to either question is `yes' yes for the standard $1/r^2$ force, then it is also yes for a any attractive `power law' $1/r^a$ force between masses, any $a > 0$.

Consider the usual three-body problem with Newtonian $1/r^2$ force between masses. Let the three masses start off at rest, and not collinear. Then they will become collinear a finite time later by a theorem I proved some time ago. Let $t_c$ denote the first such time.

Write $r_{ij} (t)$ for the distance between mass $i$ and mass $j$ at time $t$.

Question 1. If the masses are all equal and if the initial side-lengths satsify $0 < r_{12}(0) < r_{23} (0)< r_{13} (0)$
is it true that these inequalities remain in force: $0 < r_{12} (t) < r_{23} (t) < r_{13}(t)$ for $0 < t < t_c$? In other words: if the triangle starts off `scalene'' scalene (not isosceles, and of nonzero area) does it remain scalene ' up to collinearity?

Consider the usual three-body problem with Newtonian $1/r^2$ force between masses. Let the three masses start off at rest, and not collinear. Then they will become collinear a finite time later by a theorem I proved some time ago. Let $t_c$ denote the first such time.

Write $r_{ij} (t)$ for the distance between mass $i$ and mass $j$ at time $t$.

Question 1. If the masses are all equal and if the initial side-lengths satsify $0 < r_{12}(0) < r_{23} (0)< r_{13} (0)$
is it true that these inequalities remain in force: $0 < r_{12} (t) < r_{23} (t) < r_{13}(t)$ for $0 < t < t_c$? In other words: if the triangle starts off `scalene'' (not isosceles, and of nonzero area) does it remainscalene' up to collinearity?

Question 2. For general masses, is it true that the `moment of inertia' $I = m_1 m_2 r_{12} ^2 + m_2 m_3 r_{23} ^2 + m_1 m_3 r_{13} ^2$ monotonically decreases over the interval $(0, t_c)$?

Motivation: The space of collinear triangles, consisting of triangles of zero area, acts like a global Poincare section for the zero-angular momentum, negative energy three-body problem. To understand the return map'' from thissyzygy plane'' to itself the ``brake orbits''-- those solutions for which all velocities vanish at some instant -- seems to play an organizing role. Answering either questions would yield useful information about brake orbits.

Aside: I suspect that if the answers to either question is `yes' for the $1/r^2$ force, then it is also yes for a $1/r^a$ force between masses, any $a > 0$.