Liquid-drop fission energy ($Q(y)=a_S{A}^{2/3}[1-y^{2/3}-(1-y{)}^{2/3}]+a_C{Z}^2{A}^{-1/3}[1-y^{5/3}-(1-y{)}^{5/3}]$)

For two spherical fragments with the same proton fraction as the parent, let $(A_1,Z_1)=y(A,Z)$ and $(A_2,Z_2)=(1-y)(A,Z)$. In the semi-empirical mass formula without nuclear pairing, nucleon rest energies, the volume term and the nuclear symmetry energy cancel in the parent-minus-fragment mass difference. The remaining Q value is the displayed competition between increased surface energy and reduced nuclear Coulomb energy. Its symmetric value is positive whenThis is an energy-release condition, not the disappearance of a fission barrier or a prediction of a short mean lifetime.

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.