If $a^2+b^2+c^2+d^2=4$ what is the range of $a^3+b^3+c^3+d^3$? I tried to used AM-GM but could not do.Setting any one as $2$ or $-2$ and the others $0$ we get the range  as $[-8,8]$ but what is the formal way to do this?

Without resorting to higher level of math such as the use of "compactness". Observe that $a^2 \le a^2+b^2+c^2+d^2 = 4\implies a \le 2\implies a^2(a-2) \le 0$, and similarly: $b^2(b-2) \le 0, c^2(c-2) \le 0, d^2(d-2) \le 0$. Thus $a^3+b^3+c^3+d^3 \le 2(a^2+b^2+c^2+d^2) = 2\cdot 4 = 8$. And the max is $8$ which occurs when exactly one of the variables is $2$ and the others are $0$. For the min, observe that $a \ge -2, b \ge -2, c \ge -2, d \ge -2$. Thus $a^2(a+2) \ge 0, b^2(b+2) \ge 0, c^2(c+2) \ge 0, d^2(d+2) \ge 0\implies a^3+b^3+c^3+d^3 \ge -2(a^2+b^2+c^2+d^2)=-8$, which is the min and this value is achieved when exactly one of the variables is $-2$, and the others are $0$.
