Finding the maximum value

Question

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

Answer 1

$x_1=a,x_2=b,x_3=c,x_4=d,n=4,v=\sum\limits_{k=1}^nx_i^2=4$

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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 $\boxed{\forall i\in\{1,\dots,n\}, x_i \ge 0}$

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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 $\sum\limits_{k=1}^nx_i^3\le \sum\limits_{k=1}^n\sqrt{v}x_i^2=\sqrt{v}\sum\limits_{k=1}^nx_i^2\le\sqrt{v}v=v^\frac{3}{2}$

$\boxed{\sum\limits_{k=1}^nx_i^3\le v^\frac{3}{2}}$

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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 $\boxed{\exists (x_1,\dots x_n)\in\mathbb{R}^n, \sum\limits_{k=1}^nx_i^2=v \land \sum\limits_{k=1}^nx_i^3=v^{\frac{3}{2}}}$

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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.

Answer 2

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 $8$.

Answer 3

I'm guessing that the maximum value is 8, e.g. a=2, b=c=d=0, and all the cubing is "concentrated" into one number.

You could try proving by induction that if $a_1^2 + a_2^2 + ... a_n^2= c^2$, then the maximum value of $a_1^3 + a_2^3 + ... a_n^3$ is $c^3$

Answer 4

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.