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since $x_1^2+x_2^2+\cdots+x_n^2=1$ is sphere,a compact set,so $S$ has a maximal(minimal) value. But when I try to solve it using the Lagrangian multiplier method, I don't know how to solve these equations. Clearly $x_1=x_2=x_3=\cdots = x_n=\frac{1}{\sqrt{n}}$ is an extremal point, but I don't know if it's the maximal(minimal) value.

I want to know how to solve the problem. Also, could this problem be solved by elementary methods, like some inequality techniques?

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  • $\begingroup$ The point $x_1=...=x_n$ is neither the maximal nor minimal (probably for all $n$). For even $n$, choose $x_{2k}=0$ will give $S=0$ and $x_1=1/2\sqrt 2, x_2=1/\sqrt 2$ may give the maximal. $\endgroup$
    – Hhan
    Commented Feb 8, 2021 at 13:48
  • $\begingroup$ @Hhan I do not understand your maximal example, and it seems possible to have $S$ negative $\endgroup$
    – Henry
    Commented Feb 8, 2021 at 14:11
  • $\begingroup$ @Hhan It's true the point $x1=...=xn$ is neither the maximal nor minimal, but I'm not sure if it's a local maxium or minimum. $\endgroup$
    – Liuyang Guo
    Commented Feb 8, 2021 at 15:03
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    $\begingroup$ Bordered Hessian $\endgroup$
    – Zach Teitler
    Commented Feb 8, 2021 at 16:11
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    $\begingroup$ Is your question whether $x_1= \dots = x_n$ is a local maximum, local minimum, or neither, or is your question what the true global maximum and minimum are? For the second one, I will add that the symmetry $(x_1,\dots, x_n) \to (-x_1,\dots, -x_n)$ shows the set of values of $S$ is symmetric under negation and so the minimum is minus the maximum. $\endgroup$
    – Will Sawin
    Commented Feb 10, 2021 at 15:39

2 Answers 2

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For $n=3$ we need to find $$\max_{a^2+b^2+c^2=1}(a^2b+b^2c+c^2a).$$ Indeed, let $\{|a|,|b|,|c|\}=\{x,y,z\}$, where $x\geq y\geq z\geq0$.

Thus, by Rearrangement and AM-GM we obtain: $$\sum_{cyc}a^2b\leq|a|\cdot(|a||b|)+|b|\cdot(|b||c|)+|c|\cdot(|c||a|)\leq$$ $$\leq x\cdot xy+y\cdot xz+z\cdot yz=y(x^2+xz+z^2)\leq y\left(x^2+\frac{x^2+z^2}{2}+z^2\right)=$$ $$=\frac{3}{2}y(1-y^2)=\frac{3}{2\sqrt2}\sqrt{2y^2(1-y^2)^2}\leq\frac{3}{2\sqrt2}\sqrt{\left(\frac{2y^2+2-2y^2}{3}\right)^3}=\frac{1}{\sqrt3}.$$ The equality occurs for $a=b=c=\frac{1}{\sqrt3}$, which says that we got a maximal value.

For $n=4$ we need to find $$\max_{\sum\limits_{cyc}a^2=1}\sum_{cyc}a^2b.$$ Indeed, by C-S and AM-GM we obtain:

$$\sum_{cyc}a^2b\leq\sqrt{\sum_{cyc}a^2\sum_{cyc}a^2b^2}=\sqrt{(a^2+c^2)(b^2+d^2)}\leq\frac{1}{2}(a^2+c^2+b^2+d^2)=\frac{1}{2}.$$ The equality occurs for $a=b=c=d=\frac{1}{2},$ which says that we got a maximal value.

For $n\geq5$ we can use the Lagrange Multipliers method, but it does not give nice numbers.

For example, for $n=5$ the maximum occurs, when $(x_1,x_2,x_3,x_4,x_5)||(0.79...,3.24...,3.78...,2.48...,1),$ which gives a value $0.45...$

The following inequality is also true.

Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=1$. Prove that: $$a^3b^2+b^3c^2+c^3a^2\leq\frac{1}{3\sqrt3}.$$

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  • $\begingroup$ Maximize[{x[1]^2*x[2] + x[2]^2*x[3] + x[3]^2*x[1], x[1]^2 + x[2]^2 + x[3]^2 == 1}, {x[1], x[2], x[3]}]//Simplify confirms it. $\endgroup$
    – user64494
    Commented Feb 8, 2021 at 17:40
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Math experiment done with Matematica shows nothing simple and nice. $n=3$

Maximize[{x[1]^2*x[2] + x[2]^2*x[3] + x[3]^2*x[1], 
x[1]^2 + x[2]^2 + x[3]^2 == 1}, {x[1], x[2], x[3]}]

$$\left\{\frac{\left(\frac{3152 \sqrt{3}}{53}-\frac{9403}{53 \sqrt{3}}\right)^2}{\sqrt{3}}+\left(\frac{164555}{1007 \sqrt{3}}-\frac{54516 \sqrt{3}}{1007}\right)^2 \left(\frac{3152 \sqrt{3}}{53}-\frac{9403}{53 \sqrt{3}}\right)+\frac{1}{3} \left(\frac{164555}{1007 \sqrt{3}}-\frac{54516 \sqrt{3}}{1007}\right),\left\{x(1)\to \frac{1}{\sqrt{3}},x(2)\to \frac{164555}{1007 \sqrt{3}}-\frac{54516 \sqrt{3}}{1007},x(3)\to \frac{3152 \sqrt{3}}{53}-\frac{9403}{53 \sqrt{3}}\right\}\right\} $$

$n=4$

Maximize[{x[1]^2*x[2] + x[2]^2*x[3] + x[3]^2*x[4] + x[4]^2*x[1], 
x[1]^2 + x[2]^2 + x[3]^2 + x[4]^2 == 1}, {x[1], x[2], x[3], x[4]}]

$$\left\{\frac{1}{2},\left\{x(1)\to \frac{1}{2},x(2)\to \frac{1}{2},x(3)\to \frac{1}{2},x(4)\to \frac{1}{2}\right\}\right\} $$ $n=5$

Maximize[{x[1]^2*x[2] + x[2]^2*x[3] + x[3]^2*x[4] + x[4]^2*x[5] +   x[5]^2*x[1], 
x[1]^2 + x[2]^2 + x[3]^2 + x[4]^2 + x[5]^2 == 1}, {x[1], x[2], x[3], x[4], x[5]}]

produces the output of length 27003.

The same issue with the minime, e.g.

Minimize[{x[1]^2*x[2] + x[2]^2*x[3] + x[3]^2*x[1], 
x[1]^2 + x[2]^2 + x[3]^2 == 1}, {x[1], x[2], x[3]}]

$$\left\{-\frac{\left(\frac{9403}{53 \sqrt{3}}-\frac{3152 \sqrt{3}}{53}\right)^2}{\sqrt{3}}+\left(\frac{54516 \sqrt{3}}{1007}-\frac{164555}{1007 \sqrt{3}}\right)^2 \left(\frac{9403}{53 \sqrt{3}}-\frac{3152 \sqrt{3}}{53}\right)+\frac{1}{3} \left(\frac{54516 \sqrt{3}}{1007}-\frac{164555}{1007 \sqrt{3}}\right),\left\{x(1)\to -\frac{1}{\sqrt{3}},x(2)\to \frac{54516 \sqrt{3}}{1007}-\frac{164555}{1007 \sqrt{3}},x(3)\to \frac{9403}{53 \sqrt{3}}-\frac{3152 \sqrt{3}}{53}\right\}\right\} $$

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  • $\begingroup$ What is your maximum for $S$ when $n=5$? $\endgroup$
    – Henry
    Commented Feb 8, 2021 at 15:51
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    $\begingroup$ this is just undersimplified: for example, for $n=3$ we have $x(2)\to \frac{164555}{1007 \sqrt{3}}-\frac{54516 \sqrt{3}}{1007}=\frac1{\sqrt{3}}$ $\endgroup$
    – Fedor Petrov
    Commented Feb 9, 2021 at 11:20
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    $\begingroup$ It seems likely that the maximum converges as $n\to \infty$ to the maximum to the analogous optimization problem in $\mathbb Z$, i.e. maximize $\sum_{i=-\infty}^{\infty} x_i^2 x_{i+1}$ subject to $\sum_{i=-\infty}^{\infty} x_i^2 =1$. Proving this (maybe with some estimate on the rate of convergence) would be a good solution in my opinion, even if the formulas for any individual $n$ are not so nice. $\endgroup$
    – Will Sawin
    Commented Feb 10, 2021 at 20:02
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    $\begingroup$ @WillSawin It certainly does and incredibly fast, but there is no nice answer for the infinite problem either. What Yaakov pointed out is essentially the maximum: the maximizing sequence decays doubly exponentially, so beyond $x_{12}$ it is invisible to the computer and the trivial iteration like $x_i'\sim x_i+x_{i-1}^2+2x_ix_{i+1}$ converges quickly to the maximizer too. So, for all practical purposes, Yaakov's sequence is optimal (with more precision, of course, which changes only the last digit in his comment to $8$). $\endgroup$
    – fedja
    Commented Feb 13, 2021 at 1:30
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    $\begingroup$ @WillSawin Partly yes, but not in full yet. BTW, the quick proof you have gives at most $O(n^{-3/2})$ for the error just because the minimal term is of that order and you can break the chain anywhere. Of course, that is rather short of what I claimed. I'll think more of it and (I hope) post my observations later. Now I'll just note that the Lagrange multiplier theorem gives $\lambda x_i=x_{i-1}^2+2x_ix_{i+1}$ and, since the maximum is $>0.45$, we can multiply by $x_i$, add up, and get $\lambda>1.35$. Since $x_{i+1}$ is less than $0.15$ for most $i$, you get $x_i\le x_{i-1}^2$ most of the time. $\endgroup$
    – fedja
    Commented Feb 13, 2021 at 1:58

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