# An Extension of the AM-GM Inequality

The Arithmetic Mean - Geometric Mean inequality in its simplest form tells us that

For positive $x_{1}$ and $x_{2}$,

$\frac{x_{1}+x_{2}}{2}\ge \sqrt{x_{1}x_{2}}.$

This is equivalent to saying that

For positive $x_{1}$ and $x_{2}$, such that $x_{1}+x_{2}=1$,

$x_{1}x_{2}\le \frac{1}{4}.$

The problem is to prove that

For positive $x_{1}, x_{2}, \ldots\ , x_{100}$, such that $x_{1}+x_{2}+\ldots +x_{100}=1$,

$x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} + \ldots + x_{99}x_{100} \le \frac{1}{4}.$

Proof

For positive $x_{1}, x_{2}, \ldots\ , x_{100}$, such that $x_{1}+x_{2}+\ldots +x_{100}=1$,

$x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} + \ldots + x_{99}x_{100} \le \frac{1}{4}.$

### Proof

It is rather obvious that number 100 in the problem is likely to be random and the inequality holds for an arbitrary $n\ge 2$, if at all.

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

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

Since we already know that the inequality holds for $n=2$, mathematical induction naturally comes to mind. Let's check how the inequality looks like for $n=3$:

$x_{1}x_{2} + x_{2}x_{3} \le \frac{1}{4},$

provided $x_{1}+x_{2}+x_{3}=1$.

This problem easily reduces to the base case of $n=2$. Indeed, letting $x'_{1}=x_{1}+x_{3}$ gives, first of all, $x'_{1}+x_{2}=1$ and then also

$\frac{1}{4}\ge x_{1}x_{2} + x_{2}x_{3} = (x_{1} + x_{3})x_{2} = x'_{1}+x_{2}$.

Just what is needed; encouragingly, the inequality holds for $n=3$.

Trying to employ similar grouping of two terms for $n=4$ we run into a little trouble:

$x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} = x_{1}x_{2} + (x_{2} + x_{4})x_{3}$.

We do get $2$ terms as required for $n=3$ but the factors do not form a three term sequence. What is needed is to have factors $x_{2}$ in the first term and $(x_{2} + x_{4})$ equal. But surely they are not. What actually works is replacing $x_{2}$ with $(x_{2} + x_{4})$. True, these will increase the whole sum:

$x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} = x_{1}x_{2} + (x_{2} + x_{4})x_{3} \lt x_{1}(x_{2} + x_{4}) + (x_{2} + x_{4})x_{3}$.

However, on the write we do have the sum corresponding to three terms: $x_{1} + (x_{2} + x_{4}) + x_{3} = 1$. So that from already tackled case of $n=3$ we conclude that

$x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} = x_{1}x_{2} + (x_{2} + x_{4})x_{3} \lt x_{1}(x_{2} + x_{4}) + (x_{2} + x_{4})x_{3} \le \frac{1}{4}$.

Which not only proves the case of $n=4$ but also shows how to handle a general inductive step.

So assume that, provided $x_{1}+x_{2}+\ldots +x_{k}=1$,

$x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} + \ldots + x_{k-1}x_{k} \le \frac{1}{4}.$

Under this assumption we show that, for $x_{1},x_{2},\ldots, x_{k+1}$, with $x_{1}+x_{2}+\ldots +x_{k+1}=1$,

$x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} + \ldots + x_{k-2}x_{k-1} + x_{k-1}x_{k} + x_{k}x_{k+1} \le \frac{1}{4}.$

Since $x_{k-1}\lt (x_{k-1}+x_{k+1})$, we see that

\begin{align} & x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} + \ldots + x_{k-2}x_{k-1} + x_{k-1}x_{k} + x_{k}x_{k+1} \lt \\ & x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} + \ldots + x_{k-2}(x_{k-1}+x_{k+1}) + (x_{k-1}+x_{k+1})x_{k}\le \frac{1}{4} \end{align}

because the sequence if $k$ terms satisfies $x_{1}+x_{2}+\ldots +x_{k-2} + (x_{k-1}+x_{k+1}) + x_{k}=1$ and the inductive assumption applies.

(Elsewhere there is another proof - much shorter and certainly cleverer.)