Find the minimum value of $4(a^3 + b^3 + c^3) + 15abc$ with $a + b + c = 2$.

I do not know if you had to use AM-GM but the problem is quite simple using pure algebra.

Considering $$ P = 4(a^3 + b^3 + c^3) + 15abc \qquad \text{with} \qquad a+b+c=2$$ eliminate $c$ from the constaint to get $$P=3 a^2 (8-9 b)-3 a (b-2) (9 b-8)+8 (3 (b-2) b+4)$$ Now $$\frac{\partial P}{\partial a}=6 a (8-9 b)-3 (b-2) (9 b-8)=0 \implies a=\frac{2-b} 2$$ Reusing the constaint, this gives $c=a$ and then $a=b=c=\frac 23$.

Plug in $P$ and get the result.