<s>[INST] Finding the maximum value If $a^2+b^2+c^2+d^2=4$ are real numbers, then how to find the maximum value for:$$a^3+b^3+c^3+d^3$$
 [/INST] $x_1=a,x_2=b,x_3=c,x_4=d,n=4,v=\sum\limits_{k=1}^nx_i^2=4$

$\sum\limits_{k=1}^nx_i^3\le\sum\limits_{k=1}^n\left|x_i^3\right|=\sum\limits_{k=1}^n\left|x_i\right|^3$ so we can assume $\boxed{\forall i\in\{1,\dots,n\}, x_i \ge 0}$

Since $v=\sum\limits_{k=1}^nx_i^2$, $\forall i\in\{1,\dots,n\},x_i^2\le v$ that is $x_i \le \sqrt{v}$
So $\sum\limits_{k=1}^nx_i^3\le \sum\limits_{k=1}^n\sqrt{v}x_i^2=\sqrt{v}\sum\limits_{k=1}^nx_i^2\le\sqrt{v}v=v^\frac{3}{2}$
$\boxed{\sum\limits_{k=1}^nx_i^3\le v^\frac{3}{2}}$

If we take take $x_1=\sqrt{v}$ and $\forall i \in\{2,\dots,n\}, x_i = 0$
$\sum\limits_{k=1}^nx_i^2=\left(\sqrt{v}\right)^2=v$
$\sum\limits_{k=1}^nx_i^3=\left(\sqrt{v}\right)^3=v^{\frac{3}{2}}$
So $\boxed{\exists (x_1,\dots x_n)\in\mathbb{R}^n, \sum\limits_{k=1}^nx_i^2=v \land \sum\limits_{k=1}^nx_i^3=v^{\frac{3}{2}}}$

So since you know the sum is less than $v^{\frac{3}{2}}$ and this value is attained for some $ (x_1,\dots x_n)$, you can conclude that $v^{\frac{3}{2}}$ is the maximum of $\sum\limits_{k=1}^nx_i^3$.
And as you can see, the number of variables and the sum of the squares do not influence that result.
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