Which way do I prove that $36\le 4\sum_{cyc} a^3-\sum_{cyc}a^4 \le 48$? It is given that $a+b+c+d=6$ and $a^2+b^2+c^2+d^2=12$

Note that $(a-1)^4=a^4-4a^3+6a^2-4a+1$. On cyclically summing and using the data given, we have that $\sum_{cyc} (a-1)^4=\sum_{cyc}a^4-4\sum_{cyc}a^3+52$. Hence we have to prove that $$36\le 52- \sum_{cyc}(a-1)^4\le 48 \iff 4\le \sum_{cyc} (a-1)^4 \le 16$$ Now, we have that $\sum_{cyc} (a-1)^2=(\sum_{cyc} a^2)- 2(\sum_{cyc}a)+4=4$. Hence, the right inequality is clear (just note that $16=4^2=(\sum_{cyc} (a-1)^2)^2$ will have some other non negative terms other than the desired ones.) The left inequality follows from Power-mean/QM-AM/cauchy-schwartz: $$\sum_{cyc}(a-1)^4 \ge \frac{\left(\sum_{cyc} (a-1)^2\right)^2}{4}=4$$
