Proving inequality by induction only Suppose $a_1,\dots,a_n$ are real positive numbers s.t. $\prod_{i=1}^n a_i=1$. My book claims that by induction only (i.e. the use of AM-GM is forbidden), one can prove that $$\sum a_i\ge n$$ and that equiality exists if and only if $\forall i,a_i=1$. I tried to prove it:
for $n=2,$ denote $a_1=a,a_2=\frac 1 a$ then: $$0\le \frac{(a-1)^2}{a}=a+\frac{1}{a}-2\Rightarrow a_1+a_2\ge 2.$$ 
Now we need to prove it for $n+1$. Denote by $b_n=a_na_{n+1}$, By hypothesis $$a_1\cdots a_{n-1} b_n=1\Rightarrow a_1+\dots a_{n-1}+b_n\ge n$$ which means $$a_1+\dots+a_{n-1}+a_n +a_{n+1} \ge n-b_n+a_n+a_{n+1}.$$ That means I need to prove $$a_n+a_{n+1}-a_na_{n+1}\ge 1$$ but I don't know how.
How can I complete the proof?

$$a_n+a_{n+1}-a_na_{n+1}\ge 1 \Leftrightarrow 1 - a_n -a_{n+1} + a_na_{n+1} \le 0. $$
Factorize the LHS and you will see which $a_n$ and $a_{n+1}$ you should pick.
