# Michael Rozenberg's Inequality in Three Variables with Constraints

### Problem

### Solution 1

The essential fact of the proof is to observe that $2\le ab+bc+ca\le 3.\,$ Indeed,

$3(ab+bc+ca)\le (a+b+c)^2\,\Rightarrow\,ab+bc+ca\le 3.$

Let $a=\max\{a,b,c\}.\,$ Then $a\in [1,2],\,$ so that

$\begin{align} ab+bc+ca&=a(b+c)+bc=a(3-a)+bc\\ &\ge 2+bc\ge 2. \end{align}$

After usual algebraic manipulations, the required inequality is equivalent to

(*)

$\displaystyle 12\sum_{cycl}a^3\ge5\left(\sum_{cycl}a^2\right)^2-5\sum_{cycl}a^2b^2+6abc.$

From the above, $ab+bc+ca=3(1-t),\,$ with $t\in\left[0,\frac{1}{3}\right].\,$ With this in mind,

$\begin{align} &a^3+b^3+c^3=3(9t+abc)\\ &a^2+b^2+c^2=3(1+2t)\\ &a^2b^2+b^2c^2+c^2a^2=9(1-t)^2-6abc. \end{align}$

Thus, (*) can be rewritten as

$36(9t+abc)\ge 45(1+2t)^2-45(1-t)^2+36abc,$

or, equivalently, $36t\ge 5(1+2t)^2-5(1-t)^2,\,$ i.e., $2t\ge 25t^2,\,$ which is true because $t\in\left[0,\frac{1}{3}\right].$

### Solution 2

Rewriting

$\left(a^2 +b^2 +c^2 -3 a b+3 a c-3 b c\right) \left(a^2+b^2+c^2-a b-\text{ac}-b c\right)\leq 0$

By Schur's inequality: $\left(a^2+b^2+c^2-a b-\text{ac}-b c\right)\geq 0$. So we need to prove that

$g=\left( \underbrace{(a^2 +b^2 +c^2)}_{\lambda_1}-3\underbrace{(a b+ a c+ b c)}_{\lambda_2} \right) \leq 0.$

$\lambda_1$ is convex, $\lambda_2$ is concave. So the maximum of $\lambda_1=5$ is reached (under the constraint that $a+b+c=3$) at the maximum dispersion, that is $\{2,1,0\}$. The same for the minimum of $\lambda_2=6$. Hence $g\leq -1$.

### Acknowledgment

I am grateful to Leo Giugiuc for kindly communicating to me the above problem by Michael Rozenberg (Tel-Aviv, Israel), along with his solution. Solution 2 is by N. N. Taleb.

- A Cyclic But Not Symmetric Inequality in Four Variables
- An Inequality with Constraint
- An Inequality with Constraints II
- An Inequality with Constraint III
- An Inequality with Constraint IV
- An Inequality with Constraint VII
- An Inequality with Constraint VIII
- An Inequality with Constraint IX
- An Inequality with Constraint X
- Problem 11804 from the AMM
- Sladjan Stankovik's Inequality With Constraint
- An Inequality with Constraint XII
- An Inequality with Constraint XIV
- An Inequality with Constraint XVII
- An Inequality with Constraint in Four Variables II
- An Inequality with Constraint in Four Variables III
- An Inequality with Constraint in Four Variables V
- An Inequality with Constraint in Four Variables VI
- A Cyclic Inequality in Three Variables with Constraint
- Dorin Marghidanu's Cyclic Inequality with Constraint
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints II
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints III
- Inequality with Constraint from Dan Sitaru's Math Phenomenon
- Another Problem from the 2016 Danubius Contest
- Gireaux's Theorem
- An Inequality with a Parameter and a Constraint
- Unsolved Problem from Crux Solved
- An Inequality With Six Variables and Constraints
- Cubes Constrained
- Dorin Marghidanu's Inequality with Constraint
- Dan Sitaru's Integral Inequality with Powers of a Function
- Michael Rozenberg's Inequality in Three Variables with Constraints
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints IV
- An Inequality with Arbitrary Roots
- Leo Giugiuc's Inequality with Constraint
- Problem From the 2016 IMO Shortlist
- Dan Sitaru's Cyclic Inequality with a Constraint and Cube Roots
- Dan Sitaru's Cyclic Inequality with a Constraint and Cube Roots II
- A Simplified Version of Leo Giugiuc's Inequality from the AMM
- Kunihiko Chikaya's Inequality $\displaystyle \small{\left(\frac{(a^{10}-b^{10})(b^{10}-c^{10})(c^{10}-a^{10})}{(a^{9}+b^{9})(b^{9}+c^{9})(c^{9}+a^{9})}\ge\frac{125}{3}[(a-b)^3+(b-c)^3+(c-a)^3]\right)}$
- A Cyclic Inequality on [-1,1] $\left(xy+yz+zx\ge 1\right)$
- An Inequality with Two Triples of Variables $\left(\displaystyle\sum_{cycl}ux\ge\sqrt{\left(\sum_{cycl}xy\right)\left(2\sum_{cycl}uv-\sum_{cycl}u^2\right)}\right)$
- 6th European Mathematical Cup (2017), Junior Problem 4 $\left(x^3 - (y^2 + yz + z^2)x + y^2z + yz^2 \le 3\sqrt{3}\right)$
- Dorin Marghidanu's Example $\displaystyle\frac{\displaystyle\frac{1}{b_1}+\frac{2}{b_2}+\frac{3}{b_3}}{1+2+3}\ge\frac{1+2+3}{b_1+2b_2+3b_3},$
- A Trigonometric Inequality with Ordered Triple of Variables $\left((x+y)\sin x+(x-z)\sin y\lt (y+z)\sin x\right)$
- Three Variables, Three Constraints, Two Inequalities (Only One to Prove) - by Leo Giugiuc $\bigg(a+b+c=0$ and $a^2+b^2+c^2\ge 2.$ Prove that $abc\ge 0\bigg)$

|Contact| |Up| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny63036590 |