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.)

Inequalities with the Sum of Variables as a Constraint

An Inequality for Grade 8 $\left(\displaystyle\frac{1-x_1}{1+x_1}\cdot\frac{1-x_2}{1+x_2}\cdot\ldots\cdot\frac{1-x_n}{1+x_n}\ge\frac{1}{3}\right)$
An Inequality with Constraint $((x+1)(y+1)(z+1)\ge 4xyz)$
An Inequality with Constraints II $\left(\displaystyle abc+\frac{2}{ab+bc+ca}=p+\frac{2}{q}\ge q-2+\frac{2}{q}\right)$
An Inequality with Constraint V $\left(\displaystyle\prod_{k=1}^{n}x_k^{1/x_k}\le \frac{1}{n^{n^2}}\right)$
An Inequality with Constraint VI $\left(\displaystyle\prod_{k=1}^{n}\frac{1+x_k}{x_k}\ge \prod_{k=1}^{n}\frac{n-x_k}{1-x_k}\right)$
An Inequality with Constraint XI $(\sqrt{5a+4}+\sqrt{5b+4}+\sqrt{5c+4} \ge 7)$
Monthly Problem 11199 $\left(\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{25}{1+48abc}\right)$
Problem 11804 from the AMM $(10|x^3 + y^3 + z^3 - 1| \le 9|x^5 + y^5 + z^5 - 1|)$
Sladjan Stankovik's Inequality With Constraint $\left(abc+bcd+cda+dab-abcd\le\displaystyle \frac{27}{16}\right)$
Sladjan Stankovik's Inequality With Constraint II $(a^4+b^4+c^4+d^2+4abcd\ge 8)$
An Inequality with Constraint V $\left(\displaystyle\prod_{k=1}^{n}x_k^{1/x_k}\le \frac{1}{n^{n^2}}\right)$
An Inequality with Constraint VI $\left(\displaystyle\prod_{k=1}^{n}\frac{1+x_k}{x_k}\ge \prod_{k=1}^{n}\frac{n-x_k}{1-x_k}\right)$
An Inequality with Constraint XII $(abcd\ge ab+bc+cd+da+ac+bd-5)$
An Inequality with Constraint XIII $((3a-bc)(3b-ca)(3c-ab)\le 8a^2b^2c^2)$
Inequalities with Constraint XV and XVI $\left(\displaystyle\frac{a^2}{\sqrt{b^2+4}}+\frac{b^2}{\sqrt{c^2+4}}+\frac{c^2}{\sqrt{a^2+4}}\gt\frac{3}{5}\right)$ and $\left(\displaystyle\frac{a^2}{\sqrt{b^4+4}}+\frac{b^2}{\sqrt{c^4+4}}+\frac{c^2}{\sqrt{a^4+4}}\gt\frac{3}{5}\right)$
An Inequality with Constraint XVII $(a^3+b^3+c^3\ge 0)$
An Inequality with Constraint in Four Variables $\left(\displaystyle\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b}\ge\frac{1}{8}\right)$
An Inequality with Constraint in Four Variables II $(a^3+b^3+c^3+d^3 + 6abcd \ge 10)$
An Inequality with Constraint in Four Variables III $\left(\displaystyle\small{abcd+\frac{15}{2(ab+ac+ad+bc+bd+cd)}\ge\frac{9}{a^2+b^2+c^2+d^2}}\right)$
An Inequality with Constraint in Four Variables IV $\left(\displaystyle 27+3(abc+bcd+cda+dab)\ge\sum_{cycl}a^3+54\sqrt{abcd}\right)$
Inequality with Constraint from Dan Sitaru's Math Phenomenon $\left(\displaystyle b+2a+20\ge 2\sum_{cycl}\frac{a^2+ab+b^2}{a+b}\ge b+2c+20\right)$
An Inequality with a Parameter and a Constraint $\left(\displaystyle a^4+b^4+c^4+\lambda abc\le\frac{\lambda +1}{27}\right)$
Cyclic Inequality with Square Roots And Absolute Values $\left(\displaystyle \prod_{cycl}\left(\sqrt{a-a^2}+\frac{1}{2\sqrt{2}}|3a-1|\right)\ge\frac{1}{6\sqrt{6}}\prod_{cycl}\left(\sqrt{a}+\frac{1}{\sqrt{3}}\right)\right)$
From Six Variables to Four - It's All the Same $\left(\displaystyle \frac{5}{2}\le a^2+b^2+c^2+d^2\le 5\right)$
Michael Rozenberg's Inequality in Three Variables with Constraints $\left(\displaystyle 4\sum_{cycl}ab(a^2+b^2)\ge\sum_{cycl}a^4+5\sum_{cycl}a^2b^2+2abc\sum_{cycl}a\right)$
Michael Rozenberg's Inequality in Two Variables $\left(\displaystyle \sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6\right)$
Dan Sitaru's Cyclic Inequality in Three Variables II $\left(\displaystyle \sum_{cycl}\sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}\geq \frac{9}{12-2(ab+bc+ca)}+3\right)$
Dan Sitaru's Cyclic Inequality in Three Variables IV $\left(\displaystyle \sum_{cycl}\frac{(x+y)z}{\sqrt{4x^2+xy+4y^2}}\le 2\right)$
Dan Sitaru's Cyclic Inequality in Three Variables VI $\left(\displaystyle \sum_{cycl}\left[\sqrt{a(a+2b)}+\sqrt{b(b+2a)}\,\right]\le 6\sqrt{3}\right)$
An Inequality with Arbitrary Roots $\left(\displaystyle \sum_{cycl}\left(\sqrt[n]{a+\sqrt[n]{a}}+\sqrt[n]{a-\sqrt[n]{a}}\right)\lt 18\right)$
Inequality 101 from the Cyclic Inequalities Marathon $\left(\displaystyle \sum_{cycl}\frac{c^5}{(a+1)(b+1)}\ge\frac{1}{144}\right)$
Sladjan Stankovik's Inequality With Constraint II $(a^4+b^4+c^4+d^2+4abcd\ge 8)$
An Inequality with Constraint in Four Variables $\left(\displaystyle\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b}\ge\frac{1}{8}\right)$
An Inequality with Constraint in Four Variables IV $\left(\displaystyle 27+3(abc+bcd+cda+dab)\ge\sum_{cycl}a^3+54\sqrt{abcd}\right)$
Cyclic Inequality with Square Roots And Absolute Values $\left(\displaystyle \prod_{cycl}\left(\sqrt{a-a^2}+\frac{1}{2\sqrt{2}}|3a-1|\right)\ge\frac{1}{6\sqrt{6}}\prod_{cycl}\left(\sqrt{a}+\frac{1}{\sqrt{3}}\right)\right)$
From Six Variables to Four - It's All the Same $\left(\displaystyle \frac{5}{2}\le a^2+b^2+c^2+d^2\le 5\right)$
Michael Rozenberg's Inequality in Two Variables $(\displaystyle \sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6)$
Dan Sitaru's Cyclic Inequality in Three Variables II $\left(\displaystyle \sum_{cycl}\sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}\geq \frac{9}{12-2(ab+bc+ca)}+3\right)$
Dorin Marghidanu's Two-Sided Inequality $\left(\displaystyle \small{64(a+bc)(b+ca)(c+ab)}\le \small{8(1-a^2)(1-b^2)(1-c^2)}\le \small{(1+a)^2(1+b)^2(1+c^2)}\right)$
Problem 6 from Dan Sitaru's Algebraic Phenomenon $(x\sqrt{y+1}+y\sqrt{z+1}+z\sqrt{x+1}\le 2\sqrt{3})$
A Warmup Inequality from Vasile Cirtoaje $\left(a^4b^4+b^4c^4+c^4a^4\le 3\right)$
An Extension of the AM-GM Inequality
- An Extension of the AM-GM Inequality: A second look $\left(x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} + \ldots + x_{n-1}x_{n} \le \frac{a^2}{4}\right)$
Distance Inequality $\left(a^2+b^2+c^2\le\displaystyle\frac{9}{2}\right)$
Kunihiko Chikaya's Inequality with a Constraint $\left(4a^3+9b^3+36c^3\ge 1\right)$ An Inequality with Five Variables, Only Three Cyclic $\left(\displaystyle \left(a+\frac{b}{c}\right)^4+ \left(a+\frac{b}{d}\right)^4+ \left(a+\frac{b}{e}\right)^4\ge 3(a+3b)^4\right)$
Second Pair of Twin Inequalities: Twin 1 $\left(\displaystyle \prod_{i=1}^n\left(\frac{1}{a_i^2}-1\right)\ge (n^2-1)^n\right)$
Second Pair of Twin Inequalities: Twin 2 $\left(\displaystyle \prod_{i=1}^n\left(\frac{1}{a_i}+1\right)\ge (n+1)^{n}\right)$
Cyclic Inequality In Three Variables from the 2018 Romanian Olympiad, Grade 9 $\left(\displaystyle \frac{a-1}{b+1}+\frac{b-1}{c+1}+\frac{c-1}{a+1}\ge 0\right)$
Dan Sitaru's Cyclic Inequality in Three Variables IX $\left(\displaystyle \sum_{cycl}\sqrt{(x+y+1)(y+z+1)}\le 6+\sum_{cycl}\frac{x^3+y^3}{x^2+y^2}\right)$
Vasile Cirtoaje's Cyclic Inequality with Three Variables $\left(\displaystyle \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge 2\right)$ Leo Giugiuc and Vasile Cirtoaje's Cyclic Inequality $\left(\displaystyle \sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}+\sqrt{\frac{d}{1-d}}\ge 2\right)$

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