# About an inequality that looks obvious.

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.

• That is false: see $1,2,2$ and $1,4,1$. – Mindlack Feb 20 at 17:35
• @Mindlack you are right, I guess it will need more conditons. Do you think $max(a_i) \geq max(b_i)$ willl work? – StAKmod Feb 20 at 17:38
• No, same counterexample (reverse them of course). You could try and look at this, though: en.m.wikipedia.org/wiki/Karamata's_inequality – Mindlack Feb 20 at 17:41
• @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. – StAKmod Feb 20 at 17:41
• @Mindlack I am going to check this, but I think same counter example will not work. – StAKmod Feb 20 at 17:42