Proving $(1+a_1)(1+a_2)...(1+a_n) \geq 2^n$

Question

$a_1,a_2,...,a_n$ are positve numbers such that $a_1a_2...a_n=1$, prove using the AM-GM Inequality that $(1+a_1)(1+a_2)...(1+a_n) \geq 2^n$.

I managed to prove this using induction and some messing around, but I didn't use the AM-GM Inequality. I can't figure out how to prove this using the inequality, so I'd like some help. Thanks in advance to all the helpers.

Also: https://math.stackexchange.com/q/1911849/42969, https://math.stackexchange.com/a/2232876/42969 – found with Approach0 — Martin R, Oct 30 '19 at 13:38

score 1 · Accepted Answer · answered Oct 30 '19 at 13:36

1

Use that $$(1+a_1)(1+a_2)\cdot…\cdot (1+a_n)\geq 2^n\sqrt[n]{a_1\cdot a_2\cdot a_3\cdot…\cdot a_n}=2^n$$ since $$\prod_{i=1}^na_i=1$$ and by AM-GM.

answered Oct 30 '19 at 13:36

Dr. Sonnhard Graubner

97,058

3

As you wrote here: https://math.stackexchange.com/a/1911955/42969. – Martin R Oct 30 '19 at 13:37

Proving $(1+a_1)(1+a_2)...(1+a_n) \geq 2^n$

1 Answers1