<s>[INST] 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?
 [/INST] 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!
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