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Being given an integer $n\ge 2$, and $x_{i},y_{i},z_{i}\in \mathbb{R}$ ($i=1,2,\cdots,n$) such that $$\sum_{i=1}^{n}(x^3_{i}+y^3_{i}+z^3_{i})=3n$$ show that $$\sum_{i+j+k=n}x_{i}y_{j}z_{k}\le n^2.$$

I know $a^3+b^3+c^3\ge 3abc$ if $a+b+c\ge 0$.

Michael Rozenberg
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math110
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    Do you mean $x_i y_j z_k$? Or $x_i, y_i, z_i \ge 0$? Otherwise $x_i$ can be arbitrarily large and the statement is obviously wrong. – Cave Johnson Oct 27 '16 at 08:51
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    Do you have the constraint $x_i, y_i, z_i \ge 0$? – Crostul Oct 27 '16 at 09:02
  • sorry, Now I have edit, it's $x_{i}y_{j}z_{k}$,and this are real numbers – math110 Oct 27 '16 at 09:03
  • Does the verticle bar in $\sum_{n|i+j+k}x_{i}y_{j}z_{k}\le n^2$ mean "=" ? – rrogers Nov 01 '16 at 18:02
  • @Andreas, is it not that $i+j+k$ is a multiple of $n$ ? – andre Nov 05 '16 at 15:56
  • @communnites : $n=2$ is not possible or the trivial case and $x_{n-1}$,$x_{n}$,$y_{n-1}$,$y_{n}$,$z_{n-1}$,$z_{n}$ are not part of the second sum. Is this correct, no mistake in the question ? – user90369 Nov 05 '16 at 19:01
  • @rrogers In standard notation, $n|i+j+k$ would mean "$n$ divides $i+j+k$", i.e. $i+j+k$ is an integer multiple of $n$. – Andreas Nov 07 '16 at 08:20
  • @andre You're right, I wrote this again to avoid confusion. – Andreas Nov 07 '16 at 08:22
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    @communnites Yor headline asks for the sum with condition $n|i+j+k$, and the main question body asks for the sum with condition $n=i+j+k$. Please make this consistent. – Andreas Nov 07 '16 at 08:49

1 Answers1

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If you assume $x_i\geq 0$ start with $3x_i y_j z_k \leq x_i^3+ y_j^3+ z_k^3$.

Sum this over the triples $i,j,k$. You get $$3 \textrm{LHS}\leq \sum_i x_i^3\sum_{j+k=n-i} 1 + \textrm{sums for }y_i,z_i.$$

The key idea is the combinatorics here: counting how many times the term $x_i^3$ arises. The conclusion is that $$3 \textrm{LHS}\leq \sum_i (n-i-1)(x_i^3+ y_i^3+ z_i^3)$$.

Now apply the hypothesis and you should be done.

Lior Silberman
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