I'm trying to do this exercise:
Let $n \in \Bbb{N}$. Let $x_1,x_2,\ldots,x_n$ be positive real numbers such that $\prod_{i=1}^n x_i = 1$. Prove that $\sum_{i=1}^n x_i \ge n$
Here is my attempt:
Let $S = \{n \in \Bbb{N} | \sum_{i=1}^n x_i < n \}$ and $S \ne \emptyset$.
By the well ordering principle, S has a least element.
If $ n = 1$, we have:
$\prod_{i=1}^2 x_i = 1 \implies x_2 = \frac{1}{x_1} \implies \sum_{i=1}^2 x_i = \frac{x_1^2+1}{x_1}$
and $\frac{x_1^2+1}{x_1} \ge 2 \iff x_1^2+1 \ge 2x_1 \iff x_1^2 -2x_1 +1 \ge 0 \iff (x_1 - 1)^2 \ge 0$
Therefore 1 is not the least element of S.
Let k be the least element of S. We can conclude the following:
$k \ne 1 \implies k \ge 2 \implies (k-1) \in \Bbb{N} $ (1)
$\sum_{i=1}^k x_i < k \implies \sum_{i=1}^{k-1} x_i < k - x_k$ (2)
If $\prod_{i=1}^k x_i = 1$ there are 3 cases:
(i). $\prod_{i=1}^{k-1} x_i = 1$ and $x_k = 1$
(ii). $\prod_{i=1}^{k-1} x_i < 1$ and $x_k > 1$
(iii). $\prod_{i=1}^{k-1} x_i > 1$ and $x_k < 1$
Assuming cases (i) or (ii) would imply a contradiction with out assumption that k is the least element because we would have:
$\sum_{i=1}^k x_i < k \implies \sum_{i=1}^{k-1} x_i < k - x_k \le k-1$
So $(k-1) \in S$ and $S = \emptyset$
In case (iii) I'm stuck. Can we derive a contradiction assuming that $\prod_{i=1}^{k-1} x_i > 1$ and $x_k < 1$ ??? Or is there another approach to this problem? Any help will be appreciated.
