If AM-GM fails for some list of $n_0\ge 2$ numbers, then for all $n\ge n_0$, there's a list of $n$ numbers such that AM-GM fails

Question

the exercise

I want to prove this by induction, but I'm stuck.

In the induction step, I assume that there's a list of $n$ numbers such that the inequality fails, and then I should find a list of $n+1$ numbers such that the inequality fails, but i'm kind of stuck. Could I get a hint on how to continue?

Answer 1

Let $\epsilon>0$ be arbitrary. Suppose $(a_1\cdots a_{n_0})<1$.Then $(a_1\cdots a_{n_0})^{1/n_0}<(a_1\cdots a_{n_o})^{1/n_1} $ \begin{align*} \frac{a_1+a_2+\ldots+a_{n_0}+\epsilon}{n_1} &<\frac{a_1+a_2+\ldots+a_{n_0}+\epsilon}{n_0}\\ &<(a_1\cdots a_{n_0})^{1/n_0}+\frac{\epsilon}{n_0}\\ &\leq (a_1\cdots a_{n_0})^{1/n_1} \end{align*} by choosing $\frac{\epsilon}{n_0}< (a_1\cdots a_{n_0})^{1/n_1}-(a_1\cdots a_{n_0})^{1/n_0}$.

If $(a_1\cdots a_{n_0})\geq 1$, we can muyltiply the seqeunce $a_1,a_2,\ldots,a_{n_0}$ by some small enough constant $c$ where $0<c<1$ such that $(ca_1\cdots ca_{n_0})<1$. Then repeat the arugment from above.