1
$\begingroup$

I read that the rearrangement inequality deals with sorted sequences of real numbers. We have $-5>-6$ and $3>2$ , hence by rearrangement inequality we have $-15>-12$ which is obviously false.

What am I missing out?

$\endgroup$
2
  • 1
    $\begingroup$ Try reading the wikipedia page (en.wikipedia.org/wiki/Rearrangement_inequality): the claim you do has little to do with the rearragement inequality. $\endgroup$ – vrugtehagel Sep 25 '16 at 18:23
  • 1
    $\begingroup$ Rearrangement here assures $(-5)\times(3)+ (-6)\times (2)\geqslant (-5)\times (2)+(-6)\times(3)$ which is obviously true. $\endgroup$ – Macavity Sep 25 '16 at 18:27
0
$\begingroup$

In this context the rearrangement inequality says that $$ x_1 = -6 < -5 = x_2 \text{ and } y_1 = 2 < 3 = y_2 $$ implies that $$ -28 = -18 -10 = x_1 y_2 + x_2 y_1 \le x_1 y_1 + x_2 y_2 = -12 -15 =-27, $$ which is true. It does not say what you're asserting.

$\endgroup$
1
  • $\begingroup$ Oh my bad! Silly question to ask! Thanks anyway :) $\endgroup$ – user369582 Sep 25 '16 at 18:28
0
$\begingroup$

Rearrangement inequality does not have to do with multiplying inequalities by parts.

You can multiply inequalities by parts only if you know that all four sides are strictly positive. Otherwise, you will get things like: $-1<1$, $-6<3$ "thus" $6<3$ which is obviously absurd.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy