proof of arithmetic-geometric-harmonic means inequality

For the Arithmetic Geometric Inequality, I claim it is enough to prove that if${\prod }_{i=1}^n{x}_i=1$ with $x_i\ge 0$ then ${\sum }_{i=1}^n{x}_i\ge n$. The arithmetic geometric inequality for $y_1,\mathrm{\dots },y_n$ 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 $x_i\ge 0$ then ${\sum }_{i=1}^n{x}_i\ge n$ by induction on $n$.

Clear for $n=1$.

Induction Step: By reordering indices we may assume the $x_i$ are increasing, so $x_n\ge 1\ge x_1$. Assuming the statement is true for $n-1$, we have$x_2+\mathrm{\cdots }+x_{n-1}+x_1{x}_n\ge n-1$. Then,

$$\sum _{i=1}^n{x}_i\ge n-1+x_n+x_1-x_1{x}_n$$

by adding $x_1+x_n$ to both sides and subtracting $x_1{x}_n$. And so,

${\displaystyle \sum _{i=1}^n}{x}_i$$\ge n+(x_n-1)+(x_1-x_1{x}_n)$

$=n+(x_n-1)-x_1(x_n-1)$

$=n+(x_n-1)(1-x_1)$

$\ge n$

The last line follows since $x_n\ge 1\ge x_1$.