Use induction to prove that when $n≥1$, $a_i\in R$, $a_i > 0$, if $a_1a_2...a_n=1$, then $(1+a_1)(1+a_2)...(1+a_n)≥2^n$

Question

Use induction to prove that when $n≥1$, $a_i\in R$, $a_i > 0$, if $a_1a_2\cdots a_n=1$, then $(1+a_1)(1+a_2)\cdots(1+a_n)≥2^n$.*

The base case is simple, but I do not know how to complete the inductive step:

If $a_1a_2\cdots a_{n+1}=1$, then $(1+a_1)(1+a_2)\cdots(1+a_{n+1})≥2^{n+1}$.

I made a case when $a_{n+1}=1$, but could not complete the other cases.

There are other proofs of this inequality, but I do not think they use the "if" part require in this proof.

*Another way to write this inequality is as follows: $$ \left(\frac{1+a_1}{2}\right)\cdots \left(\frac{1+a_n}{2}\right) \ge 1 .$$

Answer 1

I would use AM-GM, then $$1+a_1\geq2\sqrt{a_1}$$ and $$1+a_2\geq 2\sqrt{a_2}$$ … and $$1+a_n\geq 2\sqrt{a_n}$$ so §$$(1+a_1)(1+a_2)\cdot ...\cdot(1+a_n)\geq 2^n\sqrt{a_1\cdot a_2\cdots a_n}=2^n$$

Answer 2

One of the $a_i$ is $\ge1$ and one is $\le 1$. Say $a_n\ge1\ge a_{n+1}$. Then $$(1+a_n)(1+a_{n+1})-2(1+a_na_{n+1})=(a_n-1)(1-a_{n+1})\ge0.$$ Then $$(1+a_1)\cdots(1+a_{n+1})\ge2(1+a_1)\cdots(1+a_{n-1})(1+a_na_{n+1})$$ and now apply the inductive hypothesis.