Proof with induction

Question

Let $n\in\mathbb{N}$ and $x_1,x_2,x_3.....,x_n \in\mathbb{Q}$ with $x_i > 0$ and $\prod_{i=1}^{n} x_i = 1$

prove that $\sum_{i=1}^{n}x_i \geq n$ (hint:use induction).

Been stuck on that for hours. It seems as if the terminology of the question is not well defiened because they use "Let $n\in\mathbb{N}$" and not "For each $n\in\mathbb{N}$"

Could I get some insight and help please?

Answer 1

Assume it works with $k$ terms; it certainly does when $k=1$. Consider $k+1$ terms, and permute them so $x_1$ is the greatest and $x_2$ the least. Since by hypothesis the $x_i$ have geometric mean $1$, $x_1\ge 1\ge x_2$. Now consider a sum of $k$ terms, $x_1x_2+\sum_{i=3}^{k+1}x_i\ge k$ by the inductive hypothesis. On the other hand, since $(x_1-1)(1-x_2)$ is a product of non-negative terms we have $$(x_1-1)(1-x_2)=x_1+x_2-x_1x_2-1\ge 0\implies x_1+x_2-x_1x_2\ge 1.$$Finally, $$\sum_{i=1}^{k+1}x_i=(x_1+x_2-x_1x_2)+(x_1x_2+\sum_{i=3}^{k+1}x_i)\ge 1+k$$as required.