Too generalized?

Thanks for the mentioning. I didn’t know this :)

If so, you can not pick non-negative integers $a,b$ satisfying $a+b=n$ and $b-8a=0$ , thus the maximum in such cases can't be $1/3$ .

We want to maximize $f(\mathbf{x}) \; = \; \sum_{j=1}^n \sin x_j$ subject to the constraint $g(\mathbf{x}) \; = \; \sum_{j=1}^n \sin^3x_j \; = \; 0$ For simplicity, suppose that $n \ge 9$ . Then the case $\sin x_1 = -1$ , $\sin x_2 = \sin x_3 = \cdots = \sin x_9 = \tfrac12$ , with $\sin x_j = 0$ for $j > 9$ , shows that the maximum possible value of $f$ subject to this constraint will be positive.

Since there is no boundary to the domain for $\mathbf{x}$ , the maximum of $f$ subject to the constraint $g=0$ will be identified by the method of Lagrange multipliers, so there exists $\lambda$ such that $\nabla f - \lambda \nabla g = 0$ , and hence $\cos x_j\big(1 - 3\lambda \sin^2x_j\big) \; = \; 0 \hspace{2cm} 1 \le j \le n$ If $\cos x_j = 0$ for all $j$ , then $\sin x_j = \pm1$ for all $j$ , which would imply that $f = 0$ . Thus, for a positive value of $f$ , there must be a number $0 < \mu < 1$ such that $\sin x_j \in \{1,-1,\mu,-\mu\}$ for all $j$ . Thus there must exist nonnegative integers $a,b,c,d$ such that $\begin{aligned} a + b + c + d & = \; n \\ a - b + \mu^3(c - d) & = \; 0 \\ a - b + \mu(c - d) & = \; f \end{aligned}$ (so that $a$ of the $\sin x_j$ are equal to $1$ , $b$ are equal to $-1$ , $c$ are equal to $\mu$ and $d$ are equal to $-\mu$ ). Since $f > 0$ we deduce that $c > d$ , and hence that $b > a$ . Moreover $b - a < c - d$ , and we have $\mu \; = \; \left(\frac{b-a}{c-d}\right)^{\frac13} \hspace{2cm} f \; = \; (b-a)^{\frac13}(c-d)^{\frac23} - (b-a)$ For a given choice of $a,b$ , it is clear that $f$ will be maximized by maximizing $c-d$ , so we must have $c = n-a-b$ and $d=0$ , and hence $f \; = \; (b - a)^{\frac13}(n - a - b)^{\frac23} - (b-a)$ Putting $t = b - a$ , we see that $0 \le t \le n-2a$ and that $f \; = \; t^{\frac13}(n-2a-t)^{\frac23} - t$ From this it is clear that, for a given choice of $t$ , $f$ is maximized by minimizing $a$ , and so $a=0$ . Thus we are picking $a=d=0$ and $b + c = n$ where $b < c$ , and so we want to maximize $f \; = \; b^{\frac13}(n-b)^{\frac23} - b \hspace{2cm} 1 \le b < \tfrac12n$ The function $F(u) \; = \; u^{\frac13}(1-u)^{\frac23} - u \hspace{2cm} 0 < u < 1$ has its maximum value of $\tfrac13$ at $u = \tfrac19$ , and is increasing for $u < \tfrac19$ and decreasing for $u > \tfrac19$ . Thus we deduce that $f$ will be equal to either $F\big(\lfloor \tfrac19n\rfloor \big)$ or else $F\big(\lfloor \tfrac19n\rfloor +1\big)$ , whichever is larger.

Thus the maximum value of $n^{-1}\sum_{j=1}^ny_j$ subject to the constraint $\sum_{j=1}^n y_j^3 = 0$ and $|y_j| \le 1$ for all $1 \le j \le n$ is $\tfrac{1}{n}\big[ b^{\frac13}(n-b)^{\frac23} - b \big]$ where $b$ is either $\lfloor \tfrac19n\rfloor$ or $\lfloor \tfrac19n\rfloor + 1$ , and this value is achieved by choosing $b$ of the $y_j$ equal to $-1$ , with the remaining $n-b$ values of $y_j$ equal to $\big(\tfrac{b}{n-b}\big)^{\frac13}$ . This maximum value is less than or equal to $\tfrac13$ , being equal to $\tfrac13$ when $9$ divides $n$ . I have not investigated which of the two possible values of $b$ need to be chosen for any other particular value of $n$ , but in many cases we choose $b$ to the integer closest to $\tfrac19n$ .

@Calvin Lin