2. [7] Suppose that $a, b, c$, and $d$ are real numbers such that $a+b+c+d=8$. Compute the minimum possible value of
$$
20\left(a^{2}+b^{2}+c^{2}+d^{2}\right)-\sum_{\text {sym }} a^{3} b,
$$
where the sum is over all 12 symmetric terms.
Answer:
112
Solution: Observe that
$$
\sum_{\text {sym }} a^{3} b=\sum_{\text {cyc }} a \cdot \sum_{\text {cyc }} a^{3}-\sum_{\text {cyc }} a^{4}=8 \sum_{\text {cyc }} a^{3}-\sum_{\text {cyc }} a^{4},
$$
so
$$
\begin{aligned}
20 \sum_{\text {cyc }} a^{2}-\sum_{\text {sym }} a^{3} b & =\sum_{\text {cyc }} a^{4}-8 \sum_{\text {cyc }} a^{3}+20 \sum_{\text {cyc }} a^{2} \\
& =\sum_{\text {cyc }}\left(a^{4}-8 a^{3}+20 a^{2}\right) \\
& =16(8)-4(4)+\sum_{\text {cyc }}\left(a^{4}-8 a^{3}+20 a^{2}-16 a+4\right) \\
& =112+\sum_{\text {cyc }}\left(a^{2}-4 a+2\right)^{2} \\
& \geq 112 .
\end{aligned}
$$

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 $a+b+c+d=8$.