For the Arithmetic Geometric Inequality, I claim it is enough to prove that if $\prod_{i=1}^n x_i = 1$ with $x_i \geq 0$ then $\sum_{i=1}^n x_i \geq n$ . The arithmetic geometric inequality for $y_1,\ldots,y_n$ will follow by taking $x_i = \frac{y_i}{\sqrt[n]{\prod_{k=1}^n y_k}}$ . The geometric harmonic inequality follows from the arithmetic geometric by taking $x_i = \frac{1}{y_i}$ .

So, we show that if $\prod_{i=1}^n x_i = 1$ with $x_i \geq 0$ then $\sum_{i=1}^n x_i \geq n$ by induction on $n$ .

Clear for $n=1$ .

Induction Step: By reordering indices we may assume the $x_i$ are increasing, so $x_{n} \geq 1 \geq x_1$ . Assuming the statement is true for $n-1$ , we have $x_2 + \cdots + x_{n-1} + x_1x_{n} \geq n-1$ . Then, \begin{equation*} \sum_{i=1}^n x_i \geq n-1 + x_n + x_1-x_1x_n \end{equation*}by adding $x_1+x_n$ to both sides and subtracting $x_1x_n$ . And so,