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.

Inequalities to prove:

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