= Solution
Write $r=m_1/m_2>0$. The upper ball's fraction of the available <kinetic energy> after the separated <elastic collisions> is
$$
f(r)=\frac{\tfrac12m_1v_1^2}{\tfrac12(m_1+m_2)u^2}=\frac{r(3-r)^2}{(1+r)^3},\qquad
1-f(r)=\frac{(1-3r)^2}{(1+r)^3}\ge0.
$$
Thus the maximal possible fraction is unity, attained at
$$
\boxed{\frac{m_1}{m_2}=\frac13,\qquad f_{\max}=1,\qquad v_1=2u,\quad v_2=0.}
$$
The identity proves the global bound without selecting a spurious stationary point. At the optimum the upper ball receives all the initial translational <kinetic energy>. Its subsequent rise is $4H$ above the mutual-collision level. For arbitrary large mass ratios further ground encounters may occur; the unity bound is already attained in the stated two-impact sequence.
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