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?
 A: The constraint defines a compact subset of $\mathbf{R}^4$ (a sphere), so the function to be optimized attains a minimum and maximum. Using Lagrange multipliers, we find that the minimum and maximum must be attained at points where the vectors $(a,b,c,d)$ and $(a^2,b^2,c^2,d^2)$ are collinear, hence those of $a,b,c,d$ which are nonzero must be equal. Reordering the variables, without loss of generality we may consider only the points $(1,1,1,1)$, $(2/\sqrt{3},2/\sqrt{3},2/\sqrt{3},0)$, $(\sqrt{2},\sqrt{2},0,0)$, $(2,0,0,0)$ and their opposites. The values of the second function at these points are $\pm 4$, $\pm 8/\sqrt{3}$, $\pm 4\sqrt{2}$ and $\pm 8$. Thus the maximum and minimum are $\pm 8$. Since the sphere is connected, by continuity we have that the range is $[-8,8]$.
A: 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 $0$.
A: It's enough to prove that $a^3+b^3+c^3+d^3\leq8$ 
for non-negatives $a$, $b$, $c$ and $d$ such that $a^2+b^2+c^2+d^2=4$ or
it's enough to prove that $a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}+d^{\frac{3}{2}}\leq8$ 
for non-negatives $a$, $b$, $c$ and $d$ such that $a+b+c+d=4$.
Let $f(x)=x^{\frac{3}{2}}$.
Hence, $f$ is a convex function and $(4,0,0,0)\succ(a,b,c,d)$.
Thus, by Karamata
$$\sum_{cyc}a^{\frac{3}{2}}\leq f(4)+3f(0)=8$$
and we are done!
