Find all nonnegative real numbers $a_1 \leq a_2 \leq ... \leq a_n$ satisfying
$\sum\limits_{i=1}^n a_i=12$, $\sum\limits_{i=1}^n a_i^2=18$ and $\sum\limits_{i=1}^n a_i^3=27$.
I tried it for $n=1,2,3$. For $n=1,2$ there is no solution, for $n=3$ I think that it is not possible, too.
For larger $n$ my equation system got too big. There might be an easier way or trick to find the solutions.
Thanks for your help.


$12\cdot 27=18^2$, hence we have an equality in the Cauchy-Schwarz inequality and $a_1=a_2=\ldots=a_n=\frac{12}{n}$. To have $\sum a_i^2=18$ we need $\color{red}{n=8}$, and in such a case $\sum a_i^3=27$ as wanted.

  • $\begingroup$ First of all, thanks for your quick answer. Could you explain, why 12/n for all a_i is the only solution based on the equality in the Cauchy-Schwarz inequality. $\endgroup$ – Mathefreak96 Mar 22 '18 at 0:57
  • $\begingroup$ @Mathefreak96: equality in CS occurs iff all the involved variables are equal. If they are all equal and their sum is $12$, each one of them is $12/n$. $\endgroup$ – Jack D'Aurizio Mar 22 '18 at 1:49
  • $\begingroup$ I've got the feedback, that this is not the unique solution to the problem. I searched a bit about equality in the CS-inequality and found out, that x and y not necessarily has to be linearly dependent. It's possible, that they are linearly independent in the case of degenerated bilinear forms. But how can I use this to find additional (and of course all) solutions. $\endgroup$ – Mathefreak96 Mar 28 '18 at 3:55
  • $\begingroup$ @Mathefreak96: there is no degenerate inner product here, we are dealing with he standard one. If $a_k\geq 0$ and $$\sum_{k=1}^{n}a_k \sum_{k=1}^{n}a_k^3 = \left(\sum_{k=1}^{n}a_k^2\right)^2 $$ then $a_1=a_2=\ldots=a_n$. $\endgroup$ – Jack D'Aurizio Mar 28 '18 at 4:04

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