Two Cyclic Inequalities
Introduction
Two problems discussed below have a curious story to tell. The first was proposed by the Moscow mathematician D. P. Mavlo and published in Matematica, No. 4, 1987. It was intended to be a difficult problem whose solution relied on calculus. But a short elementary solution has been submitted by a Bulgarian boy Zhivko Georgiev, a ninth-grader at the time. Zhivko's solution is given below.
The second problem has been published next year in journal Kvant (No. 11-12, 1988, problem M1136). It was dedicated to the centennial of the American Mathematical Society and was not supposed to be easy, but still elementary. The solution to the problem has been published in Kvant, No. 5, 1989. It was observed (Savchev and Andreescu, see the references at the bottom of the page) that the concise solution sounded more like a hint. But another solution can be obtained by simple substitution from the first problem. Thus it must be possible to proceed backwards and obtain another elementary solution to the latter. (Well, not that I did not try...)
Problem 1
Let $a,b,c$ be positive numbers. Prove that
$\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge\frac{3}{1+abc},$
and the equality occurs if and only if $a=b=c=1.$
Solution 1 to Problem 1
Observe that
$\begin{align}\displaystyle \frac{1}{a(1+b)}+\frac{1}{1+abc}&=\frac{1+a+ab+abc}{a(1+b)(1+abc)}\\ &=\frac{1}{1+abc}\bigg[\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}\bigg]. \end{align}$
And similarly
$\begin{align}\displaystyle \frac{1}{b(1+c)}+\frac{1}{1+abc}&=\frac{1+b+bc+abc}{b(1+c)(1+abc)}\\ &=\frac{1}{1+abc}\bigg[\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}\bigg], \end{align}$
$\begin{align}\displaystyle \frac{1}{c(1+a)}+\frac{1}{1+abc}&=\frac{1+c+ac+abc}{c(1+a)(1+abc)}\\ &=\frac{1}{1+abc}\bigg[\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a}\bigg]. \end{align}$
Before adding all three, introduce
$\displaystyle x=\frac{1+a}{a(1+b)},\space y=\frac{1+b}{b(1+c)},\space z=\frac{1+c}{c(1+a)}.$
With these, after adding up, what we get in the left-hand side
$\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc} $
whereas on the right we get
$\displaystyle \frac{1}{1+abc}\left[x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}\right]\ge \frac{6}{1+abc}, $
because for any positive $u,$ $\displaystyle u+\frac{1}{u}\ge 2,$ with equality only when $u=1.$
Putting everything together,
$\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc} \ge \frac{6}{1+abc}, $
i.e., the inequality we set out to prove.
Problem 2
Prove the inequality
$\begin{align} \displaystyle 3+(A+M+S)+\bigg(\frac{1}{A}+\frac{1}{M}+\frac{1}{S}\bigg)&+\bigg(\frac{A}{M}+\frac{M}{S}+\frac{S}{A}\bigg)\\ &\ge\frac{3(A+1)(M+1)(S+1)}{AMS+1} \end{align}$
for all positive $A,M,S.$
Solution 1 to Problem 2
First, trivially, $3+(A+M+S)=(1+A)+(1+M)+(1+S).$ Next, observe that
$\displaystyle (1+A)+\frac{1}{M}+\frac{A}{M}=\frac{(1+A)(1+M)}{M},$ etc.
So, the left-hand side transforms into
$\displaystyle\frac{(1+A)(1+M)}{M}+\frac{(1+M)(1+S)}{S}+\frac{(1+S)(1+A)}{A},$
such that after division by $(1+A)(1+M)(1+S)$ the inequality becomes
$\displaystyle\frac{1}{M(1+S)}+\frac{1}{S(1+A)}+\frac{1}{A(1+M)}\ge\frac{3}{1+AMS},$
which is the earlier inequality.
Solution 2 to Problem 2
Here's the solution published in Kvant:
The proof reduces to a rather artificial (but also ingenious!) algebraic manipulation. Multiply the difference of the left-hand and the right-hand sides by $AMS(1+AMS),$ regroup the summands, and you will obtain
$AM(M+1)(SA-1)^{2}+MS(S+1)(AM-1)^{2}+SA(A+1)(MS-1)^{2}$
a non-negative quantity. This proof was found by P. H. Diananda from the University of Singapore.
References
- S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003, pp. 101-103
Cyclic inequalities in three variables
- ab + bc + ca does not exceed aa + bb + cc
- A Cyclic Inequality in Three Variables $\left(\displaystyle\frac{a^3}{b^2(5a+2b)}+\frac{b^3}{c^2(5b+2c)}+\frac{c^3}{a^2(5c+2a)}\ge\frac{3}{7}\right)$
- A Cyclic Inequality in Three Variables II $\left(\displaystyle\frac{10a^3}{3a^2+7bc}+\frac{10b^3}{3b^2+7ca}+\frac{10c^3}{3a^2+7ab}\ge a+b+c\right)$
- A Cyclic Inequality in Three Variables III $\left(\displaystyle\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge 2\right)$
- A Cyclic Inequality in Three Variables IV $\left(\displaystyle 2\sum_{cycl}(a+b)^3+5\sum_{cycl}a^3\ge 21\sum_{cycl}a^2b\right)$
- A Cyclic Inequality in Three Variables V $\left(\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{\sqrt{3(a^2+b^2+c^2)}\cdot (a+b+c)}{ab+bc+ca}\right)$
- A Cyclic Inequality in Three Variables VI $\left(\displaystyle \frac{2(a+b+c)}{abc}\ge\sum_{cycl}\left(\sqrt{\frac{a+b}{2ac}}+\sqrt{\frac{2a}{c(a+b)}}\right)\right)$
- A Cyclic Inequality in Three Variables VII $\left(\displaystyle\sum_{cycl}x\sqrt{x^2z^2+y^4}\ge\sqrt{2}\sum_{cycl}xz\sqrt{yz}\right)$
- A Cyclic Inequality in Three Variables VIII $\left(\displaystyle\sum_{cycl}(x^2+y^2)z+\sum_{cycl}\frac{xy}{(x+y)^2}\ge 27xyz\right)$
- A Cyclic Inequality in Three Variables IX $\left(\displaystyle 9\left(\sum_{cycl}\frac{x^2}{y^2}\right)^2\ge 8\left(\sum_{cycl}\frac{x}{y}\right)\left(\sum_{cycl}\frac{x^3}{y^3}-3\right)\right)$
- A Cyclic Inequality in Three Variables X $\left(\displaystyle\sum_{cycl}\frac{1}{(a+1)^3}+4\sum_{cycl}\frac{1}{(a+1)^4}\ge\frac{9}{8}\right)$
- A Cyclic Inequality in Three Variables XI $\left(\displaystyle\sum_{cycl}\frac{1}{(a^2-ab+b^2)(b^2-bc+c^2)}\le\sum_{cycl}\frac{1}{a^4}\right)$
- A Cyclic Inequality in Three Variables XII $\left(\displaystyle\left(\sum_{cycl}\frac{1}{(a^2-ab+b^2)^6}\right)^2\le 3\sum_{cycl}\left(\frac{a+b}{a^2+b^2}\right)^{24}\right)$
- A Cyclic Inequality in Three Variables XIII $\left(\displaystyle\sum_{cycl}\frac{a^2+b^2}{a+b}+11\sum_{cycl}\frac{ab}{a+b}\gt 6\sum_{cycl}\sqrt{ab}\right)$
- A Cyclic Inequality in Three Variables XIV $\left(\displaystyle\sum_{cycl}\frac{xy}{xy+y^2+zx}\le 1\right)$
- A Cyclic Inequality in Three Variables XV $\left(\displaystyle \frac{a(a^2+b^2)}{a^3+b^3}+\frac{b(b^2+c^2)}{b^3+c^3}+\frac{c(c^2+a^2)}{c^3+a^3}\leq \sqrt{\frac{a}{b}}+\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{a}}\right)$
- A Cyclic Inequality in Three Variables XVI $\left(\displaystyle \sum_{cycl}|(a+b)(1-ab)|\lt\frac{3}{2}+\sum_{cycl}a^2+\frac{1}{2}\sum_{cycl}a^4\right)$
- A Cyclic Inequality in Three Variables XVII $\left(\displaystyle \left(\sum_{cycl}\frac{x^2}{y^2}\right)^5 \ge 9\left(\sum_{cycl}\frac{x^3}{y^2z}\right)\left(\sum_{cycl}\frac{x}{\sqrt{yz}}\right)\left(\sum_{cycl}\frac{y}{z}\right)\right)$
- A Cyclic Inequality in Three Variables XVIII $\left(\displaystyle \left(\sum_{cycl}\sqrt{ab}\right)^6 \le 27\prod_{cycl}(a^2+ab+b^2)\right)$
- A Cyclic Inequality in Three Variables XIX $\left(\displaystyle \frac{x}{(y+z)^3}+\frac{y}{(z+x)^3}+\frac{z}{(x+y)^3}\ge\frac{27}{8(x+y+z)^2}\right)$
- A Cyclic Inequality in Three Variables XX $\left(\displaystyle 5\sum_{cycl}\sqrt{ab}\le\sum_{cycl}\sqrt[4]{(a+4b)(2a+3b)(3b+2a)(4a+b)}\le 5\right)$
- A Cyclic Inequality in Three Variables XXI $\left(\displaystyle \frac{abc}{7\sqrt{7}}\le\prod_{cycl}\frac{a^2-ab+b^2}{\sqrt{a^2+5ab+b^2}}\right)$
- A Cyclic Inequality in Three Variables XXII $\left(\displaystyle \sum_{cycl}\frac{a^3}{a^2+ab+b^2}\ge\frac{a+b+c}{3}\right)$
- A Cyclic Inequality in Three Variables XXIII $\left(\displaystyle 3(a^2+b^2+c^2)^2\ge 8abc(a+b+c)+\sum_{cycl}(a^2+b^2-c^2)^2\right)$
- A Cyclic Inequality in Three Variables XXIV $\left(\displaystyle \sum_{cycl} \frac{a^2b^2 (1+a^2)(1+b^2)}{(1+a)(1+b)}\geq 4(3-2\sqrt{2})abc(a+b+c)\right)$
- A Cyclic Inequality in Three Variables XXV $\left(\displaystyle \sum_{cycl} (a-\sqrt{ab}+b)^2\cdot\sum_{cycl}(a^2-ab+b^2)^2\ge 9a^2b^2c^2\right)$
- A Cyclic Inequality in Three Variables And One More $\left(\displaystyle \left(\sum_{cycl}x^{2m+2}\right)\cdot\left(\sum_{cycl}\frac{1}{(x+y)^{2m+2}}\right)\ge\frac{9}{4^{m+1}}\right)$
- Dorin Marghidanu's Cyclic Inequality in Three Variables $\left(\displaystyle\sum_{cycl}\sqrt{a^2-ab+b^2}\sqrt{b^2-bc+c^2}\ge a^2+b^2+c^2\right)$
- 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
$\left(\displaystyle \frac{a^3}{b^2+c}+\frac{b^3}{c^2+a}+\frac{c^3}{a^2+b}\gt \frac{3}{2}(abc)^{\frac{2}{3}}\right)$
- Leo Giugiuc's Cyclic Inequality in Three Variables $\left(\displaystyle a^2+b^2+c^2\ge 3\sqrt[3]{\frac{1}{4}(a-b)^2(b-c)^2(c-a)^2}+ab+bc+ca\right)$
- Tran Hoang Nam's Cyclic Inequality in Three Variables $\left(\displaystyle\sum_{cycl}(a-b)^3(a-c)^3 \le \left(\sum_{cycl}a^2-\sum_{cycl}ab\right)^3\right)$
- Cyclic Inequality with Square Roots $\left(\displaystyle 2\sqrt{2}\sum_{cycl}xy\ge\sqrt{2xyz}\sum_{cycl}\sqrt{x}+\sum_{cycl}\sqrt{x^2z^2+y^2z^2}\right)$
- Cyclic Inequality with Logarithms $\left(\displaystyle \ln \left(a^b\cdot b^c\cdot c^a\right)+6\sum_{cycl}\frac{b(1+2a)}{1+4a+a^2}\ge 3(a+b+c)\right)$
- A Cyclic Inequality with Many Sums $\left(\displaystyle\small{ \left(\sum_{cycl}a^4\right)\left(\sum_{cycl}\frac{a}{b}\right)\left(\sum_{cycl}a^3\right)\left(\sum_{cycl}\frac{a}{c}\right)\left(\sum_{cycl}a^2\right) \ge \left(\sum_{cycl}a\right)^3\left(\sum_{cycl}\frac{1}{a}\right)^2}\right)$
- Dan Sitaru's Cyclic Inequality in Three Variables III $\left(\displaystyle 4\sqrt{\sum_{cycl}\frac{a}{(a-1)^2}}\ge\sqrt{6}(10-a-b-c)\right)$
- Imad Zak's Cyclic Inequality in Three Variables $\left(\displaystyle \frac{(a+b+c)^2}{ab+bc+ca} + \frac{ab+bc+ca}{a^2+b^2+c^2}\ge 4\right)$
- Imad Zak's Cyclic Inequality in Three Variables II $\left(\displaystyle \frac{ab+bc+ca}{(a+b+c)^2} + \frac{a^2+b^2+c^2}{ab+bc+ca}\ge \frac{4}{3}\right)$
- A Simple Cyclic Inequality in Three Variables $\left(\displaystyle 3\left(\sum_{cycl}a^2\right)^2\ge 8abc(a+b+c)+\sum_{cycl}(a^2+b^2-c^2)^2\right)$
- An Inequality with Cyclic Sums on Both Sides $\left(\displaystyle \sum_{cycl}\frac{a^9}{b^6c^2}\ge\sum_{cycl}\sqrt[6]{\frac{a^{28}}{b^{17}c^5}}\right)$
- An Inequality with Cyclic Sums on Both Sides II $\left(\displaystyle \sum_{cycl}\sqrt[6]{ab^2c^3}\ge\sum_{cycl}\sqrt[30]{a^{9}b^{10}c^{11}}\right)$
- An Inequality with Cyclic Sums on Both Sides III $\left(\displaystyle \frac{x^6z^3+y^6x^3+z^6y^3}{x^2y^2z^2}\geq \frac{x^3+y^3+z^3+3xyz}{2}\right)$
- A Cyclic Inequality in Three Variables with a Variable Hierarchy $\left(\displaystyle 2(x^2y + y^2z + z^2x+xyz) \ge (x+y) (y+z) (z+x)\right)$
- Dan Sitaru's Cyclic Inequality In Three Variables $\left(\displaystyle \frac{(5a+b)(5b+c)(5c+a)}{27(a+8c)(b+8a)(c+8b)}\geq \frac{8abc}{(5a+4b)(5b+4c)(5c+a)}\right)$
- Dan Sitaru's Cyclic Inequality in Three Variables V $\left(\displaystyle (x+y+z)^2\le \sum_{cycl}\sqrt{(x^2+xy+y^2)(y^2+yz+z^2)}\right)$
- Dan Sitaru's Cyclic Inequality in Three Variables VII $\left(\displaystyle \frac{5x+3y+z}{5z+3y+x}+\frac{5y+3z+x}{5x+3z+y}+\frac{5z+3x+y}{5y+3x+z}\ge 3\right)$
- Problem 11867 from the American Mathematical Monthly $\displaystyle \left(\left(\frac{a^2}{a^2-ab+b^2}\right)^{\frac{1}{4}} + \left(\frac{b^2}{b^2-bc+c^2}\right)^{\frac{1}{4}} + \left(\frac{c^2}{c^2-ca+a^2}\right)^{\frac{1}{4}} \le 3\right)$
- An Inequality from the 1967 IMO Shortlist $\left(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{a^8+b^8+c^8}{a^3b^3c^3}\right)$
- Birth of an Inequality $\displaystyle\left(3(a^2+b^2+c^2)^2\ge 24abc\sqrt[3]{abc}+\sum_{cyc}(a^2+b^2-c^2)^2\right)$
- A Simple Inequality in Three Variables $\displaystyle\left(\sum_{cyc}ac\left(\frac{1}{2a+b}+\frac{1}{2c+b}\right)\le \sqrt{3(a^2+b^2+c^2)}\right)$
- Cyclic Inequality in Three Variables by Marian Cucoanes $\displaystyle\left( \prod_{cycl}(\sqrt{(a+b)(a+c)}-\sqrt{bc})\ge abc\right)$
- Hung Viet's Inequality IV $\left(\displaystyle \sum_{cycl}\frac{1}{a+5b}\ge\sum_{cycl}\frac{1}{a+2b+3c}\right)$
- Cyclic Inequality with Arctangents $\left(\displaystyle 4\sum_{cycl}ab\cdot\arctan\frac{c}{b}\le\pi\sum_{cycl}a^2\right)$
- A Cyclic Inequality with Powers 2 through 7 $\left(\displaystyle \sum_{cycl}\frac{(a^7+b^7)^3}{(a^4+b^4)(a^5+b^5)(a^6+b^6)}\ge 3a^2b^2c^2\right)$
- A Long Cyclic Inequality of Degree 4 $\left(\displaystyle 4\cdot\sum_{cycl}ab\cdot\sum_{cycl}a-\left(\sum_{cycl}a\right)^3\ge\frac{\displaystyle 3\sum_{cycl}ab\left[4\sum_{cycl}ab-\left(\sum_{cycl}a\right)^2\right]}{\displaystyle \sum_{cycl}a}\right)$
- A Cyclic Inequality from the 6th IMO, 1964 $\left(\displaystyle \sum_{cycl}a^2(b+c-a)\le 3abc\right)$
- Inequality from Math Phenomenon $\left(\displaystyle \frac{a^{2n+1}}{\sqrt{bc}}+\frac{b^{2n+1}}{\sqrt{ac}}+\frac{c^{2n+1}}{\sqrt{ab}}\ge a^{2n}+b^{2n}+c^{2n}\right)$
- A Cyclic Inequality from Math Phenomenon $\left(\displaystyle \frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}\ge\frac{12}{(a+b+c)^2}\right)$
- Trigonometric Inequality with Integrals $\left(\small{\Omega(a,b)+\Omega(b,c)+\Omega(c,a)\le \sqrt{2}(a^2+b^2+c^2),\;\Omega(a,b)=\int_a^{2a}\int_b^{2b}|\sin (x-y)\cos (x+y)-\sin (x+y)|dxdy}\right)$
- A Cyclic Inequality in Three Variables by Uche E. Okeke $\left(\displaystyle \frac{(a+b)(b+c)(c+a)}{2}\ge abc+ \frac{(ab+bc+ca)^2}{a+b+c}\right)$
- Problem 4142 From Crux Mathematicorum $\displaystyle\left(\Bigr(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\Bigr)^{\frac{(a+b+c)^2}{a^2+b^2+c^2}}\leq \Bigr(1+\frac{a}{b}\Bigr)\Bigr(1+\frac{b}{c}\Bigr)\Bigr(1+\frac{c}{a}\Bigr)\right)$
- Two-Sided Inequality by Dorin Marghidanu $\left(\displaystyle \sqrt{2}\le\sum_{cycl}\frac{b+c}{a+\sqrt{2(b^2+c^2)}}\le 2\right)$
- A Cyclic Inequality In Three Variables by Sorin Radulescu $\left(\displaystyle \left[\sum_{cycl}x(y-z)^2\right]^3\ge 54\prod_{cycl}x(y-z)^2\right)$
- Problem 1 from the 2017 Canada MO $\left(\displaystyle\left(\frac{a}{b-c}\right)^2+\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2\gt 2\right)$
- An Easy Cyclic Inequality And a Remark $\left(\displaystyle 1\le\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\le 2\right)$
- A Cyclic Inequality from India In Three Variables And More $\left(\displaystyle \sum_{cycl}\sqrt{a^4+a^2b^2+b^4}+(6-\sqrt{3})\sum_{cycl}ab\ge 2\left(\sum_{cycl}a\right)^2\right)$
- Dan Sitaru's Cyclic Inequality in Three Variables VIII $\left(\displaystyle (xy+yz+zx)\sum_{cycl}\sqrt{x^2+xy+y^2}\le 3\sqrt{\prod_{cycl}(x^2+xy+y^2)}\right)$
- Hadamard's Determinant Inequalities and Applications I $\left((2-a-b-c+abc)^2\le (a^2+2)(b^2+2)(c^2+2)\right)$
- An Inequality with Two Triples of Variables II $\left(\displaystyle\small{ ax+by+cz+\sqrt{\left(\sum_{cycl}a^2\right)\left(\sum_{cycl}x^2\right)}\ge\frac{2}{3}\left(\sum_{cycl}a\right)\left(\sum_{cycl}x\right)}\right)$
- Two Cyclic Inequalities $\left(\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge\frac{3}{1+abc}\right),$ $\left(\begin{align} \displaystyle 3+(A+M+S)+\bigg(\frac{1}{A}+\frac{1}{M}+\frac{1}{S}\bigg)&+\bigg(\frac{A}{M}+\frac{M}{S}+\frac{S}{A}\bigg)\\&\ge\frac{3(A+1)(M+1)(S+1)}{AMS+1}\end{align}\right)$
- Ji Chen's Inequality $\left(\displaystyle (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4}\right)$
- 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 $\left(\displaystyle \sum_{cycl}(a-b)\cdot\left(\frac{a}{b}\right)^n\ge \sum_{cycl}(a-b)\cdot\frac{a}{b}\right)$
- 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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