If $n$ positive real numbers $a_i~(1\le i\le n)$ satisfy $\sum\limits_{i=1}^na_i=\sum\limits_{i=1}^na_i^2=n-1$, find the maximum or supremum of
\[M=a_1^3+a_2^3+\cdots+a_n^3.\]
I think the maximum of $M$ is at $a_1=\frac{2n-2}n$, $a_2=a_3=\cdots=a_n=\frac{n-2}n$ and permutations.
For the proof, I thought of the $uvw$ theory. That is to pick three unique numbers, and adjust so that two of them are equal and the cube sum is maximized.
However, such adjustment seem to be endless, so we need a smarter way…
In addition if we can prove that it has got maximum and not only supremum, we could directly inspect the case of maximal, then prove by contraction.
Best Answer
WLOG, assume that $a_1 = \max(a_1, a_2, \cdots, a_n)$.
We have $$a_1^2 + (a_2^2 + a_3^2 + \cdots + a_n^2) \ge a_1^2 + \frac{(a_2 + a_3 + \cdots + a_n)^2}{n - 1} = a_1^2 + \frac{(n - 1 - a_1)^2}{n - 1}$$ which results in $$a_1 \le \frac{2(n - 1)}{n}. \tag{1}$$
We have $$\left(a_k - \frac{n - 2}{n}\right)^2(a_1 - a_k) \ge 0$$ which results in $$a_k^3 \le \left(2 + a_1 - \frac{4}{n}\right)a_k^2 + \left(\frac{4}{n} - 1 + \frac{4a_1}{n} - \frac{4}{n^2} - 2a_1\right)a_k + \frac{(n - 2)^2a_1}{n^2}. \tag{2}$$
Using (1) and (2), we have \begin{align*} M &\le a_1^3 + \left(2 + a_1 - \frac{4}{n}\right)\sum_{k=2}^n a_k^2 + \left(\frac{4}{n} - 1 + \frac{4a_1}{n} - \frac{4}{n^2} - 2a_1\right)\sum_{k=2}^n a_k\\[6pt] &\qquad + \frac{(n - 2)^2a_1}{n^2}\cdot (n - 1)\\[6pt] &= a_1^3 + \left(2 + a_1 - \frac{4}{n}\right)(n - 1 - a_1^2) + \left(\frac{4}{n} - 1 + \frac{4a_1}{n} - \frac{4}{n^2} - 2a_1\right)(n - 1 - a_1)\\[6pt] &\qquad + \frac{(n - 2)^2a_1}{n^2}\cdot (n - 1)\\[6pt] &= a_1 + n - 1 - \frac{4}{n} + \frac{4}{n^2}\\[6pt] &\le \frac{2(n - 1)}{n} + n - 1 - \frac{4}{n} + \frac{4}{n^2}. \end{align*}
Also, when $a_1 = \frac{2(n - 1)}{n}$ and $a_2 = a_3 = \cdots = a_n = \frac{n - 2}{n}$, we have \begin{align*} M &= \left( \frac{2(n - 1)}{n}\right)^3 + (n - 1)\left(\frac{n - 2}{n}\right)^3\\ &= \frac{2(n - 1)}{n} + n - 1 - \frac{4}{n} + \frac{4}{n^2}. \end{align*}
Thus, the maximum of $M$ is $\frac{2(n - 1)}{n} + n - 1 - \frac{4}{n} + \frac{4}{n^2}$.