A Little More of Algebra for an Inequality
A Little Less of Calculus for a Generalization
Source
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.
Cyclic inequalities in three variables
- ab + bc + ca does not exceed aa + bb + cc
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- A Cyclic Inequality in Three Variables V $\left(\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{\sqrt{3(a^2+b^2+c^2)}\cdot (a+b+c)}{ab+bc+ca}\right)$
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- A Cyclic Inequality in Three Variables XXV $\left(\displaystyle \sum_{cycl} (a-\sqrt{ab}+b)^2\cdot\sum_{cycl}(a^2-ab+b^2)^2\ge 9a^2b^2c^2\right)$
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- Dorin Marghidanu's Cyclic Inequality in Three Variables II
$\left(\displaystyle\sum_{cycl}\frac{ab}{(a+c)(b+c)} \ge\frac{3}{4}\right)$
Dorin Marghidanu's Cyclic Inequality in Three Variables III
$\left(\displaystyle \frac{a^3}{b^2+c}+\frac{b^3}{c^2+a}+\frac{c^3}{a^2+b}\gt \frac{3}{2}(abc)^{\frac{2}{3}}\right)$
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- Dan Sitaru's Cyclic Inequality in Three Variables V $\left(\displaystyle (x+y+z)^2\le \sum_{cycl}\sqrt{(x^2+xy+y^2)(y^2+yz+z^2)}\right)$
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- An Inequality from the 1967 IMO Shortlist $\left(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{a^8+b^8+c^8}{a^3b^3c^3}\right)$
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- Hung Viet's Inequality IV $\left(\displaystyle \sum_{cycl}\frac{1}{a+5b}\ge\sum_{cycl}\frac{1}{a+2b+3c}\right)$
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- Problem 4142 From Crux Mathematicorum $\displaystyle\left(\Bigr(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\Bigr)^{\frac{(a+b+c)^2}{a^2+b^2+c^2}}\leq \Bigr(1+\frac{a}{b}\Bigr)\Bigr(1+\frac{b}{c}\Bigr)\Bigr(1+\frac{c}{a}\Bigr)\right)$
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- Hadamard's Determinant Inequalities and Applications I $\left((2-a-b-c+abc)^2\le (a^2+2)(b^2+2)(c^2+2)\right)$
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- A Cyclic Inequality of Degree Four $\left(\displaystyle a^4b+b^4c+c^4a+2(a+b+c)\ge \sqrt{3}(ab+bc+ca)\right)$
- A Little More of Algebra for an Inequality, A Little Less of Calculus for a Generalization
- A Cyclic Inequality in Three Variables XXVI $\left(\displaystyle \sum_{cycl}\frac{(x+y)^4+1}{(x+y)^6+1}\le\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right)$
- Dorin Marghidanu's Inequality with Powers and Reciprocals $\left(\displaystyle \sum_{cycl}\frac{a}{a^2bc+b^4+c^4}\le\frac{1}{abc}\right)$
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