I recently came up with a proof by simple induction of the arithmetic mean - geometric mean inequality that I haven't found here. I'm sure it isn't new.
My questions:
(1) Is this correct?
(2) Is this new here?
Proof by induction of AM-GM inequality (AMGMI).
Statement.
If $a_i > 0$ for $1 \le i \le n$ then $\left(\dfrac{\sum_{i=1}^n a_i}{n}\right)^n \ge \prod_{i=1}^n a_i $ with equality only when all $a_i$ are equal.
Proof.
For $n=1$, this is $\dfrac{a_1}{1} \ge a_1$ which is true.
For $n=2$ this is $\left(\dfrac{a_1+a_2}{2}\right)^2 \ge a_1a_2 $ which is the same as $\left(\dfrac{a_1-a_2}{2}\right)^2 \ge 0 $ which is true with equality only when $a_1 = a_2$.
For the induction step, we will use Bernoulli's inequality (BI) in the form $(1+x)^n \ge 1+nx $ for integer $n \ge 1$, real $x > -1$ with equality only when $x = 0$ or $n = 1$.
Proof of BI.
True for $n = 1$.
If $n \ge 1$, then
$\begin{array}\\ (1+x)^{n+1} &=(1+x)^n(1+x)\\ &\ge (1+nx)(1+x)\\ &= 1+(n+1)x + nx^2\\ &\ge 1+(n+1)x\\ \end{array} $
with equality only when $x = 0$.
Now the proof.
Let $S_m =\sum_{i=1}^{m} a_i $.
If AMGMI is true for $n \ge 2$ then
$\begin{array}\\ \left(\dfrac{S_{n+1}}{n+1}\right)^{n+1} &=\left(\dfrac{S_n+a_{n+1}}{n+1}\right)^{n+1}\\ &=\left(\dfrac{\dfrac{n+1}{n}S_n-\dfrac{1}{n}S_n+a_{n+1}}{n+1}\right)^{n+1}\\ &=\left(\dfrac1{n}S_n+\dfrac{-\dfrac{1}{n}S_n+a_{n+1}}{n+1}\right)^{n+1}\\ &=\left(\dfrac1{n}S_n\right)^{n+1}\left(1+\dfrac{-1+\dfrac{na_{n+1}}{S_n}}{n+1}\right)^{n+1}\\ &\ge\left(\dfrac1{n}S_n\right)^{n+1}\left(1+\left(-1+\dfrac{na_{n+1}}{S_n}\right)\right) \qquad \text{(from BI since } \dfrac{na_{n+1}}{S_n} > 0)\\ &=\left(\dfrac1{n}S_n\right)^{n+1}\left(\dfrac{na_{n+1}}{S_n}\right)\\ &=\left(\dfrac1{n}S_n\right)^{n}a_{n+1}\\ &\ge\left(\prod_{i=1}^n a_i\right)a_{n+1} \qquad\text{(induction hypothesis)}\\ &= \prod_{i=1}^{n+1} a_i\\ \end{array} $
There is equality only when all $a_i$ for $1 \le i \le n$ are equal (call the common value $a$) and $\dfrac{na_{n+1}}{S_n} = 1 $ or $a_{n+1} =\dfrac{S_n}{n} =a $.
