$a_1, ... . a_{n+1} > 0$, then $(\prod_{i=1}^{n+1}a_i)(\sum_{i=1}^{n+1} 1/a_i^n) \ge \sum_{i=1}^{n+1}a_i$?

Question

If $a_1, ... . a_{n+1} > 0$, then how to prove that $(\prod_{i=1}^{n+1}a_i)(\sum_{i=1}^{n+1} 1/a_i^n) \ge \sum_{i=1}^{n+1}a_i$ ? I couldn't get anywhere, I don't know where to start . Please help . Thanks in advance

Answer 1

Let $a_i=\frac{1}{x_i}$ and $n+1=k$. Hence, we need to prove that $$x_1^{k-1}+x_2^{k-1}+...+x_k^{k-1}\geq \sum_{i=1}^k\prod_{j\neq i}x_j,$$ which is Muirhead because $(k-1,0,...,0)\succ(1,1,...,1,0)$.

Done!

Answer 2

Writing $P=\prod_{i=1}^{n+1}a_i ; S=\sum_{i=1}^{n+1}1/a_i^n$ , we have to prove $PS \ge \sum_{i=1}^{n+1}a_i$ , i.e.

$P((n+1)S-S) \ge n \sum_{i=1}^{n+1}a_i$ i.e. $P[(S-1/a_1^n)+...+(S-1/a_{n+1}^n)]\ge n \sum_{i=1}^{n+1}a_i$ .

Now by A.M.-G.M. inequality, $\sum _{j \ne i} 1/a_j^n \ge n/\prod_{j \ne i} a_j$ i.e. $S-1/a_i^n \ge na_i/P , \forall i=1(1)n+1$ , hence multiplying both sides by $P$ and summing we get the required inequality