Given are the real numbers $x_1,x_2,\dots,x_9$ which satisfy the conditions $\sum_{i=1}^9x_i^3 = 0$ and $|x_i|\leq1$

Find the maximum value of $\sum_{i=1}^9x_i$

Intiutively the sum has its max. value when $x_1=-1$ and $x_2,x_3,\dots,x_9=1/2$ but I don't know how to prove it.


2 Answers 2


Your intuition seems right. Let $y_i = x_i^3$ and $f(t) = \sqrt[3]t$, then we have given $\sum y_i = 0$, to find the maximum of $\sum f(y_i)$. Note that $f$ is concave in $[0, 1]$ and convex in $[-1, 0]$. Hence we get

$$ f(u) + f(v) \le \begin{cases} f(-1) + f(1+u+v) & \qquad u, v \in [-1,0] \\ 2 f\left(\dfrac{u+v}2\right) & \qquad u, v \in [0, 1] \end{cases} \tag{A}$$

Also, we have, if $ u < 0 < v$: $$ f(u) + f(v) \le \begin{cases} f(-1) + f(1+u+v) & \qquad u+v \le 0 \\ f(0) + f(u+v) & \qquad u+v \ge 0 \end{cases} \tag{B}$$

WLOG, we can assume $y_i$ are in ascending order. Let the first $k$ terms be negative. Then using the reasoning in $(A)$ above we have $$\sum_{i=1}^k f(y_i) \le (k-1)f(-1) + f\left(\sum_{i=1}^k (y_i +k-1) \right)$$ Thus it is clear that we can set $k-1$ of these variables to $-1$ for maximising the above partial sum. Let the remaining negative variable be $y_k$.

Further, for the terms which are $\ge 0$, we have from $(A)$: $$\sum_{i=k+1}^9 f(y_i) \le (9-k)f\left(\frac1{9-k} \sum_{i=k+1}^9 y_i \right) $$ hence we can replace all non-negative numbers with their arithmetic mean $\mu$ to maximise this partial sum.

Now we have $y_k < 0 < \mu $ and hence using $(B)$ we can replace $(\mu, y_k)$ with $(-1, 1+\mu+y_k)$ or $(0, \mu+y_k)$ to increase the sum (depending on the sign of $\mu + y_k$), thereby making all negative terms equal to $-1$. So we have that the maximum is when some $k$ of the $y_i$s are $-1$, and the remaining are all equal. Obviously $\mu = \dfrac{k}{9-k}$, giving a maximum sum of $$g(k) = kf(-1) + (9-k)f\left(\frac{k}{9-k}\right) = \sqrt[3]{k(9-k)^2}-k $$

It is easy to check that $k = 1$ gives the maximum among integers in $[1, 4]$, and so the maximum value is when $y_1= -1$ and $y_2 = ... y_9 = \frac18$ or when any one of the $x_i$ is $-1$ and the rest are all $\frac12$

P.S. Very similar to another problem : Maximum of the sum of cube


Search with Lagrange in (ie, with the restrictions):

$$\sum x_i^3=0;$$

$$\sum x_i^3=0,\qquad x_k=1, \qquad{\rm for each}\ k=1,\cdots, 9;$$

$$\sum x_i^3=0,\qquad x_k=-1,\qquad{\rm for each}\ k=1,\cdots, 9;$$

$$\sum x_i^3=0,\qquad x_j=1,x_k=1 \qquad{\rm for each}\ j\ne k=1,\cdots, 9;$$


Awfully long.

  • $\begingroup$ Is there a more elementary way? $\endgroup$ Jan 28, 2014 at 16:25
  • $\begingroup$ I can not think of another. And is worse... see my edit. $\endgroup$ Jan 28, 2014 at 16:27

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