# A Little More of Algebra for an Inequality A Little Less of Calculus for a Generalization

### Problem

Let $a,b,c$ be positive real numbers, $n$ a positive integer. Prove that

$\displaystyle \sum_{cycl}(a-b)\cdot\left(\frac{a}{b}\right)^n\ge \sum_{cycl}(a-b)\cdot\frac{a}{b}.$

### Solution 1

Define $\displaystyle U_n= \sum_{cycl}(a-b)\cdot\left(\frac{a}{b}\right)^n.$ The task is to prove that $U_n\ge U_1.$ We shall prove that for any $n\ge 1,$ $U_{n+1}-U_n\ge 0,$ from which the claim will follow step by step.

\displaystyle\begin{align} U_{n+1}-U_n&=\sum_{cycl}(a-b)\cdot\left(\frac{a}{b}\right)^{n+1}-\sum_{cycl}(a-b)\cdot\left(\frac{a}{b}\right)^n\\ &=\sum_{cycl}(a-b)\cdot\left(\frac{a}{b}\right)^n\left(1-\frac{a}{b}\right)\\ &=\sum_{cycl}\frac{(a-b)^2}{b}\cdot\left(\frac{a}{b}\right)^n\ge 0. \end{align}

We thus have, $U_{n}\ge U_{n-1}\ge\ldots U_2\ge U_1.$

### Solution 2

The inequality can be rewritten as

$\displaystyle \sum_{cycl}\frac{a^{n+1}}{b^n}-\sum_{cycl}\frac{a^n}{b^{n-1}}\ge\sum_{cycl}\frac{a^2}{b}-\sum_{cycl}a.$

Define $\displaystyle T_n=\sum_{cycl}\frac{a^{n+1}}{b^n}.$ The inequality can be rewritten as $T_n-T_{n-1}\ge T_1-T_0,$ for $n\ge 1.$

Now, from $a^2+b^2\ge 2ab,$ $\displaystyle \frac{a^2}{b}\ge 2a-b,$ and similarly, $\displaystyle \frac{b^2}{c}\ge 2b-c,$ $\displaystyle \frac{c^2}{a}\ge 2c-a.$ Adding up the three gives $T_1-T_0\ge 0.$

Multiplying the three by $\displaystyle \frac{a^{n-1}}{b^{n-1}},$ $\displaystyle \frac{b^{n-1}}{c^{n-1}},$ $\displaystyle \frac{c^{n-1}}{a^{n-1}},$ respectively, we get

\displaystyle \begin{align} \frac{a^{n+1}}{b^n}\ge 2\frac{a^n}{b^{n-1}}-\frac{a^{n-1}}{b_{n-2}}\\ \frac{b^{n+1}}{c^n}\ge 2\frac{b^n}{c^{n-1}}-\frac{b^{n-1}}{c_{n-2}}\\ \frac{c^{n+1}}{a^n}\ge 2\frac{c^n}{a^{n-1}}-\frac{c^{n-1}}{a_{n-2}}, \end{align}

adding which yields $T_n\ge 2T_{n-1}-T_{n-2},$ or equivalently $T_n-T_{n-1}\ge T_{n-1}+T_{n-2}.$ As before we can descent to $T_1-T_0:$

$T_n-T_{n-1}\ge T_{n-1}+T_{n-2}\ge\ldots\ge T_1-T_0.$

### Solution 3

The inequality can be written as

$\displaystyle \sum_{cyc}(a-b)\left[\left(\frac{a}{b}\right)^n-\left(\frac{a}{b}\right)\right]\geq 0.$

The inequality is trivially satisifed for $n=1$. For $n\gt 1$, let $x=a/b$. The first term in the cyclic sum can be written as

$b(x-1)(x^n-x)=bx(x-1)^2(1+x+x^2+...+x^{n-2})\geq 0.$

Same logic applies to the other two terms and the result follows.

### Generalization

If Let $a,b,c,t$ be positive real numbers. Prove that

$\displaystyle \sum_{cycl}(a-b)\cdot\left(\frac{a}{b}\right)^{t+1}\ge \sum_{cycl}(a-b)\cdot\frac{a}{b}.$

### Proof

The inequality is equivalent to

$\displaystyle \sum_{cycl}\frac{a-b}{b^t}\cdot\frac{a}{b}\cdot (a^t-b^t)\ge 0.$ By the Mean Value Theorem, there are real numbers $u\in [a,b],$ $v\in [b,c],$ $w\in [c,a]$ such that $a^t-b^t=t(a-b)u^{t-1},$ $b^t-c^t=t(b-c)v^{t-1},$ $c^t-a^t=t(c-a)w^{t-1}.$ Thus the latter inequality reduces to

$\displaystyle \frac{t(a-b)^2u^{t-1}}{b^t}\cdot\frac{a}{b}+ \frac{t(b-c)^2v^{t-1}}{c^t}\cdot\frac{b}{c}+\frac{t(c-a)^2w^{t-1}}{a^t}\cdot\frac{c}{a}\ge 0,$

which is obviously true.

### Acknowledgment

The problem has been originally posted by Dorin Marghidanu at the CutTheKnotMath facebook page, along with Solutions 1 and 2. Leo Giugiuc commented on with a generalization (and its proof). Solution 3 is by Amit Itagi.