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If $a^2+b^2+c^2+d^2=4$ are real numbers ,then how to find the maximum value for :$$a^3+b^3+c^3+d^3$$

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Please, show you work. Thanks. – Cortizol Mar 1 '13 at 9:48
Use AM-GM inequality .:) – Inceptio Mar 1 '13 at 11:40

To maximize/minimize $a^3+b^3+c^3+d^3$ under the constraint that $a^2+b^2+c^2+d^2=4$ means we need $$ a^2\delta a+b^2\delta b+c^2\delta c+d^2\delta d=0\tag{1} $$ for any deltas so that $$ a\,\delta a+b\,\delta b+c\,\delta c+d\,\delta d=0\tag{2} $$ Standard orthogonality arguments say that there must be a $\lambda$ so that $$ (a^2,b^2,c^2,d^2)=\lambda\,(a,b,c,d)\tag{3} $$ That is, $(a^2,b^2,c^2,d^2)$ must be perpendicular to any delta which is perpendicular to $(a,b,c,d)$.

$(3)$ says that $a^2-\lambda a=0$. Thus, either $a=0$ or $a=\lambda$. The same is true for $b$, $c$, and $d$.

There are a finite number of possibilities, each depending on how many of $b$, $c$, and $d$ are zero and whether $\lambda$ is positive or negative.

$0$ zeroes: $\lambda=\pm1$ and $a^3+b^3+c^3+d^3=\pm4$

$1$ zero: $\lambda=\pm\sqrt{\frac43}$ and $a^3+b^3+c^3+d^3=\pm4\sqrt{\frac43}$

$2$ zeroes: $\lambda=\pm\sqrt{2}$ and $a^3+b^3+c^3+d^3=\pm4\sqrt{2}$

$3$ zeroes: $\lambda=\pm2$ and $a^3+b^3+c^3+d^3=\pm8$

Thus, the maximum of $a^3+b^3+c^3+d^3$ is $8$.

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$\sum\limits_{k=1}^nx_i^3\le\sum\limits_{k=1}^n\left|x_i^3\right|=\sum\limits_{k=1}^n\left|x_i\right|^3$ 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 $\forall i \in\{2,\dots,n\}, x_i = 0$



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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Why not just knowing that the maximum for $a,b,c,d =2$ then $a^3 <= 2a^2$? – user64505 Mar 1 '13 at 11:19
Yeah right. Don't need $x_m$. – xavierm02 Mar 1 '13 at 12:00
Nice proof (+1). You forced me to resort to calculus, which may be too much for the algebra-precalculus tag. – robjohn Mar 1 '13 at 14:20

Put $a^2=:p\geq0$, $\ b^2=:q\geq0$, $\ c^2=:r\geq0$, $\ d^2=:s\geq0$, and note that $$(p+q)^{3/2}-p^{3/2}={3\over2}\int_p^{p+q} x^{1/2}\ dx\geq {3\over2}\int_0^q x^{1/2}\ dx=q^{3\over2}\ .$$ It follows that $p^{3/2}+q^{3\over2}\leq (p+q)^{3/2}$, so that we get $$\eqalign{a^3+b^3+c^3+d^3 &\leq p^{3/2}+q^{3/2}+r^{3/2}+s^{3/2}\leq (p+q)^{3/2}+(r+s)^{3/2}\cr &\leq (p+q+r+s)^{3/2}= 4^{3/2}=8\ .\cr}$$ Putting $a=2$, $b=c=d=0$ the maximal possible value $8$ is indeed realized.

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I think you wrote 2/3 where you meant 3/2. – xavierm02 Mar 1 '13 at 16:14
@xavierm02: Thank you. I have edited that. – Christian Blatter Mar 1 '13 at 16:19
Hadn't spotted those ones >_< I was talking about the powers of $q$ line 2 and 3. – xavierm02 Mar 1 '13 at 16:25

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