# An Inequality with Constraint XII

### Statement

### Solution 1

Denote $a-1=t,\;$ $1-b=z,\;$ $1-c=y\;$ and $1-d=x.\;$ Then

$1\ge x\ge y\ge z\ge 0\;$ and $x+y+z=t.$

The required inequality translates into

$-xyzt-xyz+t(xy+yz+zx)\ge 0$

which is equivalent to

$(x+y+z)(xy+yz+zx)\ge xyz(x+y+z+1).$

The latter inequality is true because, by the AM-GM inequality, $(x+y+z)(xy+yz+zx)\ge 9xyz\;$ and the obvious $4\ge 1+x+y+z.$

Let's study the cases of equality. If $xyz\gt 0,\;$ the inequality is strict. Assume then $xyz=0.\;$ Then either $x+y+z=0\;$ or $xy+yz+zx=0.\;$ It follows that $(x,y,z)=(\alpha,0,0),\;$ $\alpha\in [0,1].\;$ In other words, the equality is achieved only when $(a,b,c,d)=(1+\alpha,1,1,1-\alpha).$

### Solution 2

Define $f(a,b,c,d)=5+abcd-(ab+bc+cd+da+ac+bd).\;$ Then

$f(a,b,c,d)-f(a+b-1,1,c,d)=(1-cd)(a-1)(1-b)\ge 0\\ \begin{align} f(a+b-1,1,c,d)&=(a+b-2)(a+b+cd-3)\\ &=(a+b-2)(1-c)(1-d)\\ &\ge 0. \end{align}$

### Acknowledgment

The problem and Solution 1 are by Leo Giugiuc who also communicated to me Solution 2 from the Art of Problem Solving forum.

- 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 $\displaystyle \small{\left(\frac{(a^{10}-b^{10})(b^{10}-c^{10})(c^{10}-a^{10})}{(a^{9}+b^{9})(b^{9}+c^{9})(c^{9}+a^{9})}\ge\frac{125}{3}[(a-b)^3+(b-c)^3+(c-a)^3]\right)}$

|Contact| |Front page| |Contents| |Algebra| |Store|

Copyright © 1996-2017 Alexander Bogomolny62316791 |