 Symmetric fission maximizes the positive liquid-drop energy release

Ignore nuclear pairing and use the continuous proportional-fragment model of liquid-drop fission energy. Put $S=a_S{A}^{2/3}$, $x=a_C{Z}^2/(a_{S}A)$ and $t=y^{1/3}+(1-y{)}^{1/3}$. As $y$ runs from zero to $1/2$, $t$ rises from $1$ to $2^{2/3}$. The identity $u^3+v^3=1$ with $u=y^{1/3}$ and $v=(1-y{)}^{1/3}$ givesThe derivative changes sign at most once, from negative to positive. Thus the global maximum on the closed continuous interval occurs at either the no-split endpoint $Q=0$ or the symmetric split. If positive fission energy is available, the symmetric value is the global maximum. This qualification matters: symmetry alone makes $y=1/2$ stationary but does not make it a maximum for all parameters. In fact, $Q^{\prime \prime }(1/2)<0$ requires $Z^2/A>(2/5)a_S/a_C$; below that value the symmetric split is a local minimum. Real discrete fragments and nuclear-shell corrections are outside this continuous smooth model.