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Let $x_{i}\in \mathbb R$ $(i=1,2,\dots,n)$, $S_{k}=x^k_{1}+x^k_{2}+\cdots+x^k_{n}$, and $S_{2}=1$. Find the maximum and minimum of $S_{1}S_{3}$.

If $x_{i}>0$, then using Cauchy-Schwarz inequality we have $$S_{1}S_{3}\ge (S_{2})^2=1.$$

But for real numbers $x_{i}$ it seems hard, and I have note $-1\le x_{i}\le 1$. So $S_{1}S_{3}$ exist maximum and minmum of the value when $n=2$, since $S_{2}=x^2_{1}+x^2_{2}=1$, then let $x_{1}=\cos{a},x_{2}=\sin{a}$. So $$S_{1}S_{3}=(\sin{a}+\cos{a})(\sin^3{a}+\cos^3{a})=(1+2\sin{a}\cos{a})(1-\sin{a}\cos{a})=(1+\sin{(2a)})(1-\dfrac{1}{2}\sin{(2a)})=(1+t)(1-\dfrac{1}{2}t)=-\dfrac{1}{2}(t-\frac{1}{2})^2+\dfrac{9}{8}\in [0,\dfrac{9}{8}]$$ where $t=\sin{(2a)}$, and when $t=-1$ and $t=\frac{1}{2}$ take the minimum and maximum of the value.

user26857
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math110
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  • Have you tried for $ n = 2, 3$? What are the results there? – Calvin Lin Mar 31 '22 at 15:40
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    Using Cauchy-Schwarz to obtain $S_{1}S_{3}\ge (S_{2})^2=1$ is not restricted to positive reals. – Andreas Apr 02 '22 at 15:20
  • @Andreas I think you are mistaken about that. Consider the example $x_1=\frac{1}{2},x_2=\frac{-1-\sqrt{5}}{4},x_3=\frac{-1+\sqrt{5}}{4}$. Then $S_2=1$ but $S_1=0$, so $S_1S_3 \geq S_2^2$ does not hold. – Ewan Delanoy Apr 03 '22 at 11:09
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    @EwanDelanoy Good point - let's digest it. What OP really does is to use Cauchy-Schwarz for roots (and not for the $x_i$, $x_i^3$) as follows: $$ Q = \sum (\pm|x_i|^{\frac12})^2 \sum (\pm|x_i|^{\frac32})^2 \underbrace{\ge}_{CS} (\sum |x_i|^{\frac12} |x_i|^{\frac32} )^2 = (\sum x_i^{2} )^2 = S_2^2 $$ So it's the $\pm|x_i|^{\frac12} $ and $ \pm|x_i|^{\frac32} $ which are not restricted to positive reals (as CS is generally not). However, $Q \ne S_1 S_3$ in general, other than for positive ${x_i}$. Since only then, $x_i = (\pm|x_i|^{\frac12})^2 $ and $x_i^3 = (\pm|x_i|^{\frac32})^2 $. – Andreas Apr 03 '22 at 16:32

1 Answers1

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Define a mean $m$ with $n m = \sum_{i=1}^{n}x_{i}$. Define deviations $d_i = x_i -m$, where obviously $\sum_{i=1}^{n}d_{i} = 0$. Since $1 = \sum_{i=1}^{n}x_{i}^2 = n m^2 + \sum_{i=1}^{n}d_{i}^2$ we have $|m| < 1/\sqrt n$. Let's continue working with fixed $m$ in the following, and come back to the discussion of $m$ later.

Now write the target function $$ F = \sum_{i=1}^{n}x_{i}\sum_{i=1}^{n}x^3_{i} = nm (nm^3 + 3 m \sum_{i=1}^{n}d_{i}^2 + \sum_{i=1}^{n}d_{i}^3) = n^2m^4 + 3 n m^2(1-nm^2) + n m \sum_{i=1}^{n}d_{i}^3 $$ For the last term, we would like to use some known result. Let $d_i = \sqrt{1 - n m^2} a_i$, then $\sum_{i=1}^{n}a_{i}^2 = 1$ and $$ F = n^2m^4 + 3 n m^2(1-nm^2) + n m (1 - nm^2)^{3/2} \sum_{i=1}^{n}a_{i}^3 $$ So for bounding the last term, the question arises about the maximum and minimum of $\sum_{i=1}^{n}a_{i}^3$ under the constraints $\sum_{i=1}^{n}a_{i} = 0$ and $\sum_{i=1}^{n}a_{i}^2 = 1$. This has been solved for example here, showing that $$ - \frac{n-2}{\sqrt{n(n-1)}} \le \sum_ia_i^3 \le \frac{n-2}{\sqrt{n(n-1)}}. $$

Building this into the target function, we get that the maximum (for some $m$) is achieved for $m$ and $\sum_ia_i^3$ having the same sign, giving $$ F_{max}(|m|) = n^2m^4 + 3 n m^2(1-nm^2) + n |m| (1 - nm^2)^{3/2} \frac{n-2}{\sqrt{n(n-1)}} $$ Conversely, the minimum is achieved for $m$ and $\sum_ia_i^3$ having opposite sign, giving $$ F_{min}(|m|) = n^2m^4 + 3 n m^2(1-nm^2) - n |m| (1 - nm^2)^{3/2} \frac{n-2}{\sqrt{n(n-1)}} $$

Since $n m^2 \le 1$, one can write $nm^2 = \cos(x)^2$ and bring this into the form $$ F(x) = \cos(x)^4 + 3 \cos(x)^2\sin(x)^2 \pm \cos(x) \sin(x)^3 \frac{n-2}{\sqrt{n-1}} \quad \tag{1} $$ however this does not appear to provide easier solutions, so it was not followed further on. It is useful for a large-$n$ discussion, see bottom.

This leaves us with the discussion of $m$. We have that the global $ F_{max} = {\max}_{|m|}{F_{max}(|m|)} $ and likewise the global $ F_{min} = {\min}_{|m|}{F_{min}(|m|)} $.

For $n=2$, we have in the range $0 \le |m| \le 1/\sqrt{2}$ that $F_{max} = {\max}_{|m|}4m^4 + 6 m^2(1-2m^2) = 9/8 $ at $|m|≈0.61 < 1/\sqrt 2 $ and $F_{min} = {\min}_{|m|}4m^4 + 6 m^2(1-2m^2) = 0 $ at $|m|=0$ , as was already stated in the question.

For general $n$, closed form results are not directly obtainable. For example, for $n=3$ we have (using WolframAlpha) that
$ F_{max} = {\max}_{|m|}9 m^4 + 9 m^2 (1 - 3 m^2) + \sqrt{3/2} |m| (1 - 3 m^2)^{3/2} ≈1.22307$ at $|m|≈0.46416 < 1/\sqrt 3$ and $ F_{min} = {\min}_{|m|}9 m^4 + 9 m^2 (1 - 3 m^2) - \sqrt{3/2} |m| (1 - 3 m^2)^{3/2} ≈-0.040397$ at $|m|≈0.065262 < 1/\sqrt 3$

For $n=4$, WolframAlpha provides $ F_{max}≈1.30667$ at $|m|≈0.38302$ and $F_{min}≈-0.102869$ at $|m|≈0.086824$.

One observes that with rising $n$, maxima get higher and minima lower.

Indeed, when observing eq. (1), for large $n$ the first two terms have no relation to $n$ and stay of order $1$, whereas the last term grows with roughly $\propto \sqrt{n}$. Since the maximum of $ \cos(x) \sin(x)^3$ is $3 \sqrt{3}/16 \simeq 0.325$, we have that $F_{max} {{n \; {\text{large}}} \atop {\to}} \simeq 0.325 \sqrt{n}$ and $F_{min} {{n \; {\text{large}}} \atop {\to}} \simeq - 0.325 \sqrt{n}$.

Andreas
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