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: The inequality you need to prove (I think you should change the $n-1$ to $n+1$ if you follow what you have on the previous line),
$$
a_n+a_{n+1}-a_na_{n+1}\geq 1
$$
is not in general true. Take $a_n=a_{n+1}=2$ (which is perfectly valid). This gives zero in the left-hand side.
So, maybe what you can do, is to assume that you, in your product $b_n=a_na_{n+1}$ choose the elements $a_n$ and $a_{n+1}$ that has a certain property? 
A: $$a_n+a_{n+1}-a_na_{n+1}\ge 1 \Leftrightarrow 1 - a_n -a_{n+1} + a_na_{n+1} \le 0. $$
Factorize the LHS and you will see which $a_n$ and $a_{n+1}$ you should pick.
A: What we want to show is that
$\prod_{i=1}^n a_i=1
\implies \sum a_i\ge n
$.
This is certainly true for $n=1$.
Suppose it is true for $n$.
Suppose we have
$\prod_{i=1}^{n+1} a_i=1
$.
We want to show that
$ \sum_{i=1}^{n+1} a_i\ge n+1
$.
As a first try,
let's look at
$\prod_{i=1}^{n} a_i
$.
If
$A
=(\prod_{i=1}^{n} a_i)^{1/n}
$
and
$b_i
=a_i/A
$,
then
$\prod_{i=1}^{n} b_i
= 1
$,
so
$\sum_{i=1}^{n} b_i
\ge n
$,
or
$\sum_{i=1}^{n} a_i
\ge An
$.
Therefore
$\sum_{i=1}^{n+1} a_i
\ge An+a_{n+1}
$.
If we can show that
$ An+a_{n+1}
\ge n+1
$, we are done.
This is the same as
$\frac{An+a_{n+1}}{n+1}
\ge 1
$.
But
$1
=\prod_{i=1}^{n+1} a_i
=a_{n+1}\prod_{i=1}^{n} a_i
=a_{n+1}A^n
$,
so this is the same as
$\frac{An+a_{n+1}}{n+1}
\ge a_{n+1}A^n
$
or
$An+a_{n+1}
\ge (n+1)a_{n+1}A^n
$
or
$An+1/A^n
\ge (n+1)a_{n+1}A^n
$
By the AM-GM means inequality,
since $nA$ is $n$ copies of $A$,
$\begin{array}\\
\frac{ An+a_{n+1}}{n+1}
&\ge \sqrt[n]{A^n a_{n+1}}\\
&= \sqrt[n]{(\prod_{i=1}^n a_i) a_{n+1}}\\
&= \sqrt[n]{\prod_{i=1}^{n+1} a_i}\\
&= 1\\
\end{array}
$
and,
to my pleasant surprise,
we are done.
(added even later)
We are not allowed to use
the AM-GM inequality.
However,
I will now show,
based on some earlier work of my own,
that this particular use
on the AM-GM inequality,
in which all but one of the values
are equal,
is a consequence of
Bernoulli's inequality.
Bernoulli's inequality,
which is easily proved by induction,
states that
$(1+x)^m \ge 1+mx
$
if $x \ge 0$ and
$m$ is a positive integer
with equality if and only if
$m = 1$
or
$x = 0$.
Therefore
$(1+\frac{v/u-1}{m})^m 
\ge 1+m\frac{v/u-1}{m} 
= \frac{v}{u}
$ with
equality only if 
$m=1$ or $v/u-1 = 0
$,
which is the same as 
$v = u$.
Multiplying by $u^m$,
this becomes
$\begin{array}\\
u^{m-1}v
&\le u^m(1+\frac{v/u-1}{m})^m \\
&= (u+\frac{v-u}{m})^m \\
&= (\frac{v+(m-1)u}{m})^m \\
\text{or}\\
(u^{m-1}v)^{1/m}
&\le \frac{v+(m-1)u}{m}\\
\end{array}
$
But this is just
the AM-GM inequality
with all but one of the values equal!
Therefore,
this use of the
AM-GM inequality
is OK to use,
since it is a consequence
of Bernoulli's inequality,
not the full
AM-GM inequality.
(added later)
If there is equality,
then
$A = a_{n+1}
$.
Since
$1 
=A^n a_{n+1}
$,
$1 = a^{n+1}_{n+1}
$,
so
$a_{n+1} = 1$.
This implies that
there is equality for
$n$,
so that
$a_n = 1$.
This allows us to
work our way back down
in an odd kind of
inverse induction
to show that
equality at any step
shows that
at that step
and all preceding steps
$a_i = 1$.
This is weird,
but I think that
it is correct.
