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My question is:

Prove or disprove: $$ a_1,a_2,...,a_n\;\text{ and }\;b_1,b_2,...,b_n\;\text{ are positive real numbers }(n\geq 2) $$ Then if $a_1a_2...a_n=b_1b_2,...,b_n$ and $\min(a_i) \leq \min(b_i),\max(a_i)\geq \max(b_i)$

We can conclude that $a_1+a_2+...+a_n \geq b_1+b_2+...+b_n$?

I used to think this as an obvious result, since the AM is greater if the sequence is more discrete when the GM is fixed.

However, today I tried to prove it, or find an counter-example, both ways failed.

The only way I tried is to raise both sides to the $n^{th}$ power, and try to factor the difference, however it does not go well.

Then I just ran out of way to deal with it. Any help or hint will be greatly appreciated.

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    $\begingroup$ That is false: see $1,2,2$ and $1,4,1$. $\endgroup$ – Mindlack Feb 20 at 17:35
  • $\begingroup$ @Mindlack you are right, I guess it will need more conditons. Do you think $max(a_i) \geq max(b_i)$ willl work? $\endgroup$ – StAKmod Feb 20 at 17:38
  • $\begingroup$ No, same counterexample (reverse them of course). You could try and look at this, though: en.m.wikipedia.org/wiki/Karamata's_inequality $\endgroup$ – Mindlack Feb 20 at 17:41
  • $\begingroup$ @Mindlack actually the original statement is $a\leq b\leq c\leq d\leq e\leq f$,and that $adf=bce$, that is the original one, sorry I should type it. $\endgroup$ – StAKmod Feb 20 at 17:41
  • $\begingroup$ @Mindlack I am going to check this, but I think same counter example will not work. $\endgroup$ – StAKmod Feb 20 at 17:42

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