, 2024

Solution:
Observe that
$$\sum_{\mathrm{sym}} a^{3} b=\sum_{\mathrm{cyc}} a \cdot \sum_{\mathrm{cyc}} a^{3}-\sum_{\mathrm{cyc}} a^{4}=8 \sum_{\mathrm{cyc}} a^{3}-\sum_{\mathrm{cyc}} a^{4}$$
so
$$\begin{aligned} 20 \sum_{\mathrm{cyc}} a^{2}-\sum_{\mathrm{sym}} a^{3} b & =\sum_{\mathrm{cyc}} a^{4}-8 \sum_{\mathrm{cyc}} a^{3}+20 \sum_{\mathrm{cyc}} a^{2} \\ & =\sum_{\mathrm{cyc}}\left(a^{4}-8 a^{3}+20 a^{2}\right) \\ & =16(8)-4(4)+\sum_{\mathrm{cyc}}\left(a^{4}-8 a^{3}+20 a^{2}-16 a+4\right) \\ & =112+\sum_{\mathrm{cyc}}\left(a^{2}-4 a+2\right)^{2} \\ & \geq 112 . \end{aligned}$$
Equality is achieved when $(a, b, c, d)=(2+\sqrt{2}, 2+\sqrt{2}, 2-\sqrt{2}, 2-\sqrt{2})$ and permutations. This can be checked by noting that for all $x \in\{a, b, c, d\}$, we have $x^{2}-4 x+2=0$, so equality holds in the final step, and for this assignment of variables we have $a+b+c+d=8$.