Theorem: Suppose there is a sequence of positive real numbers such that $a_{1}a_{2}... a_{n}=1$ then
$$(1+a_{1})(1+a_{2})...(1+a_{n})\geq2^{n}$$
(Prove by induction, do not use geometric mean)
I believe I have a proof, but I am unsure it is correct. Could you help me identify any mistakes or find a more direct approach?
Proof: Let $n=1$ then clearly $a_{1}=1$ and $(1+1)\geq2$
Assume the claim is true for all sequences of length $k<n$. Then from a sequence $a_{1}a_{2}..a_{n}$, let $c=a_{i}a_{j}$ where $a_{i}\geq1$ and $a_{j}\leq1$. We know we can pick these because otherwise the product must be less or than greater one.
Then $c(a_{1}a_{2}...a_{n})= 1$ where $i\neq j$ and $i\neq k$ and by the induction hypothesis:
$$(c+1)(a_{1}+1)(a_{2}+1)...(a_{n}+1)\geq 2^{n-1}$$
And
$$(1+a_{i})(1+a_{j})=a_{i}a_{j}+a_{i}+a_{j}+1$$
We want to show this product is greater than $2(c+1)$.
$$(1-a_{j})\geq (1-a_{j})$$
$$a_{i}(1-a_{j})\geq (1-a_{j})$$ since $a_{i}\geq 1$.
$$a_{i}-a_{i}a_{j}\geq (1-a_{j})$$
$$a_{i} + a_{j} \geq a_{i}a_{j} + 1$$
$$a_{i} + a_{j} + (a_{i}a_{j} + 1) \geq a_{i}a_{j} + 1 + (a_{i}a_{j} + 1)$$
$$(1+a_{i})(1+a_{j}) \geq 2(a_{i}a_{j} + 1) = 2(c+1)$$
Finally:
$$(1+a_{i})(1+a_{j})(a_{1}+1)...(a_{n}+1)\geq 2(c+1)\frac{2^{n-1}}{c+1}=2^{n}$$
Note: This exercise comes from Udi Manber's Intro to Algorithms.
$$ a+b \geq 2\sqrt{ab} $$
$$ (1+a_1)(1+a_2)\ldots(1+a_n) \geq 2\sqrt{a_1} \cdot 2\sqrt{a_2} \ldots 2\sqrt{a_n} = 2^n\sqrt{a_1a_2a_3 \ldots a_n} = 2^n $$