I think it's familiar, but if not, here it is.
Given $n$ positive real numbers $a_1, a_2, … a_n$:
Prove that, if: $$a_1\cdot a_2\cdot\cdot\cdot a_n=1$$
Then: $$a_1+a_2+\ldots+a_n\geq n$$
I need proof by induction, not by using the AM-GM inequality (the latter is easy).
Thanks in advance.
Best Answer
Assume that the inequality is true for $n-1$.
Without loss of generality, assume that $a_1$ and $a_n$ are respectively the maximum and minimum among $a_1, a_2, \dots, a_n$.
Note that $a_1 \ge 1$ and $a_n \le 1$
It thus follows that $$(1-a_1)(a_n-1) \ge 0 \Leftrightarrow a_1+a_n \ge a_1a_n+1$$
Note that since $a_2 \times a_3 \times \dots \times a_{n-1} \times a_1a_{n}=1$, our indutive hypotheis implies $$n \le a_2+a_3\dots+a_{n-1}+a_1a_n+1 \le a_1+a_2 +\dots+a_n$$
We are done.