# An Extension of the AM-GM Inequality: A second look

Elsewhere I proved by induction starting with the Arithmetic Mean - Geometric Mean inequality for two terms an inequality equivalent to the following - a slightly more general - one:

For positive $a,x_{1}, x_{2}, \ldots\ , x_{n}$, such that $x_{1}+x_{2}+\ldots +x_{n}=a$,

$x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} + \ldots + x_{n-1}x_{n} \le \frac{a^2}{4}.$

Here I offer a lovely, insightful proof that does not refer to the AM-GM inequality at all; it is rather based on a well known property of parabola.

### Lemma

For any $a, x$,

$x(a - x) \le \frac{a^{2}}{4}.$

### Proof

Parabola $f(x)=x(a-x)$ has two roots $x=0$ and $x=a$; midway between the two, $x=\frac{a}{2}$ serves as its axis of symmetry. This is where the function attends its maximum: $f(x)\le f(\frac{a}{2}) = \frac{a^2}{4}$.

The proof of the proposition is based on a clever observation that the terms in the sum at hand are each a product of two factors: one with an odd, the other en even index. Define

$x=x_{1}+x_{3}+x_{5}+\ldots$.

(The sum contains all the odd indices not exceeding n.) Then

$a-x=x_{2}+x_{4}+x_{6}+\ldots$.

Now, it is obvious that

$x(a-x)\ge x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} + \ldots + x_{n-1}x_{n}$

simply because the left-hand side contains all the terms on the right and many more.