= Solution
The lines of centres from either lower cylinder to the upper cylinder make $60^\circ$ with the horizontal. Let $N_L,N_R$ be the upward <normal forces> on the top cylinder from the left and right cylinders, and let $Q$ be the lower cylinders' mutual normal <force>. While all three contacts remain closed, the triangular configuration translates horizontally with common <acceleration> $a$; frictionless normals exert no <torques> about the cylinder centres.
For the top cylinder, <Newton's second law> gives
$$
\frac{N_L-N_R}{2}=ma,\qquad\frac{\sqrt3}{2}(N_L+N_R)=mg.
$$
Thus $N_L=m(g/\sqrt3+a)$ and $N_R=m(g/\sqrt3-a)$. The right lower cylinder has horizontal equation $N_R/2+Q=ma$, so
$$
Q=\frac m2\left(3a-\frac g{\sqrt3}\right).
$$
A contact may push but cannot pull: $N_L,N_R,Q\geq0$. The limiting inequalities are therefore
$$
\boxed{\frac{g}{3\sqrt3}\leq a\leq\frac g{\sqrt3}.}
$$
At the lower bound, $Q=0$ and the lower pair are just losing compression; at the upper bound, $N_R=0$ and the upper-right contact is just losing compression. All three pairwise normal <forces> are strictly positive only inside the open interval. The ground normal <forces> are positive throughout this range. The applied horizontal <force> on the left cylinder is $F=3ma$, as follows from the whole system's horizontal <momentum> balance.
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