# Prove inequality$\sum\limits_{n|i+j+k}x_{i}y_{j}z_{k}\le n^2$

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 $a+b+c\ge 0$.

• 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. Oct 27, 2016 at 8:51
• Do you have the constraint $x_i, y_i, z_i \ge 0$? Oct 27, 2016 at 9:02
• sorry, Now I have edit, it's $x_{i}y_{j}z_{k}$,and this are real numbers Oct 27, 2016 at 9:03
• Does the verticle bar in $\sum_{n|i+j+k}x_{i}y_{j}z_{k}\le n^2$ mean "=" ? Nov 1, 2016 at 18:02
• @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. Nov 7, 2016 at 8:49

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 $$3 \textrm{LHS}\leq \sum_i (n-i-1)(x_i^3+ y_i^3+ z_i^3)$$.