$36 \leq 4(a^3+b^3+c^3+d^3) - (a^4+b^4+c^4+d^4)\leq4 8.$ Let $a,b,c,d \in \Bbb R$, $a+b+c+d=6$ and $a^2+b^2+c^2+d^2=12$. Then
$$36 \leq 4(a^3+b^3+c^3+d^3) - (a^4+b^4+c^4+d^4)\leq4
8.$$
I have found only two bounds: $216 \geq a^3+b^3+c^3+d^3$ and $144 \geq a^4+b^4+c^4+d^4$.
How to prove this inequality?

For upper bound, we have
$$ x^3(4-x) \leq 4x^2\;\; \forall x$$
and we are done.

For lower bound:
Let $a+b+c =x$, then $d=6-x$ and by Cauchy we have $a^2+b^2+c^2\geq x^2/3$
so $$12  \geq {x^2\over 3} +(6-x)^2$$ so $x\in (3,6) $ and so $0\leq d\leq 3$. But this holds also for $a,b,c$.  
Idea for finish:
Let $f(x) = 4x^3-x^4$, for $x\in[0,3]$ we have $4x^3-x^4\geq x^3$, so we have by Cauchy
$$ f(a)+f(b)+f(c)\geq a^3+b^3+c^3\geq {(a^2+
b^2+c^2)^2\over a+b+c} = {(12-d^2)^2\over 6-d}$$
So we have to check if following holds:
$$ f(d)+ {(12-d^2)^2\over 6-d}\geq 36$$
