14
$\begingroup$

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 $a+b+c\ge 0$.

$\endgroup$
9
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 27, 2016 at 8:51
  • 1
    $\begingroup$ Do you have the constraint $x_i, y_i, z_i \ge 0$? $\endgroup$
    – Crostul
    Oct 27, 2016 at 9:02
  • $\begingroup$ sorry, Now I have edit, it's $x_{i}y_{j}z_{k}$,and this are real numbers $\endgroup$
    – math110
    Oct 27, 2016 at 9:03
  • $\begingroup$ Does the verticle bar in $\sum_{n|i+j+k}x_{i}y_{j}z_{k}\le n^2$ mean "=" ? $\endgroup$
    – rrogers
    Nov 1, 2016 at 18:02
  • 1
    $\begingroup$ @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. $\endgroup$
    – Andreas
    Nov 7, 2016 at 8:49

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .