Given $ n $ positive integer determine the maximum value of $ x ^ 3_1 + x ^ 3_2 + ... + x ^ 3_n $ for $ x_j $, with $ 1 \leq j \leq n $, real numbers satisfying $ x_1 + x_2 +. ... + x_n = 0 $ and $ x ^ 2_1 + x ^ 2_2 + ... + x ^ 2_n = 1 $ I think it works by using Lagrange multipliers.

But I tried to use sums of Newton: $ p (x) = (x-x_1) \cdots (x-x_n) $ Then there are things like Vieta like $ x_1 + x_2 + \cdots + x_n $ etc. But their product appears two by two, three by thre.But I can't keep going


1 Answer 1


Well, lets try using Vieta as you mention, and see what we have. Clearly $p(x)$ is monic. Further, for $n<3$ there is nothing to optimise. Else,

Coefficient of $x^{n-1}$ is $-\sum x_i=0$.

Coefficient of $x^{n-2}$ is $-\frac12(\sum x_i)^2+ \frac12\sum x_i^2=-\frac12$, and that’s all the conditions we have.

Coefficient of $x^{n-3}$ is, say, $C=\frac16(\sum x_i)^3-\frac12\sum x_i \cdot \sum x_i^2+\frac13\sum x_i^3=\frac13 \sum x_i^3$, so we seek to maximise $3C$.

So $p(x)$ is of the form: $$p(x)=x^n-\tfrac12x^{n-2}-Cx^{n-3}+\dots$$

As we don’t have any more info, we note for all real roots, we need the $(n-3)$ th derivative of $p(x)$ also to have all real roots. This derivative is $$\frac{n!}{3!}x^3-\frac{(n-2)!}2 \,x-(n-3)!C$$

Equivalently $x^3-\dfrac3{n(n-1)}x-\dfrac6{n(n-1)(n-2)}C=0 $ has three real roots. This has non-negative discriminant iff $$\Delta_3=4\left(\frac3{n(n-1)}\right)^3-27 \left(\frac6{n(n-1)(n-2)}\right)^2C^2\geqslant 0 \\ \implies \sum x_i^3=3C \leqslant \frac{n-2}{\sqrt{n(n-1)}}$$

It may be noted when one among the $x_i$ is $(n-1)\alpha$ and all others $-\alpha$, where $\alpha = \frac{1}{\sqrt{n(n-1)}}$, this maximum is indeed achieved.


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