= 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 $S=a_SA^{2/3}$, $x=a_CZ^2/(a_SA)$ 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 $v=(1-y)^{1/3}$ gives
$$
\frac QS=1+x-\frac{t^2+2/t}{3}+\frac{x}{9}(t^5-5t^2-5/t),\qquad
\frac{d(Q/S)}{dt}=\frac{t^3-1}{9t^2}[5x(t^3-1)-6].
$$
The 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, $Q''(1/2)<0$ 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.
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