To prove the given inequality of n numbers

Question

Given $ a_1 + a_2 + ... + a_n = S$ where $a_1,...,a_n$ are positive reals and all $a_i$s are not equal, then show that $$ \prod_{i=1}^n\dfrac{S-a_i}{n-1} > a_1.a_2...a_n$$ I began by considering the arithmetic mean of $n$ numbers $ (S-a_1)/(n-1),...,(S-a_n)/(n-1)$ that come out to be $S/n$ which is also the AM of $a_1,..,a_n$. How do I proceed on from here? Thank you in advance.

Answer 1

You have to use AM-GM inequality for each item on the left side. That is, for example, for the first item we have $$\frac{(S-a_1)}{n-1} >=(a_2a_3...a_n)^{\frac{1}{n-1}}$$ Then just notice that the product of all expressions on the right reduces to $$a_1a_2..a_n$$ since each $a_k$ pops $(n-1)$ times.

Answer 2

You are on the right track, but it is not $n$ numbers but $n-1$. Then use AM-GM inequality and take the product of the LHS and RHS. Now under the $n-1$ root you may have $a_i$ , $n-1 $ times....