# Determine the Maximum Value

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

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.