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.