# An Inequality with Constraint XIV

### Statement

### Solution

Case 1: $a+b+c=0.\,$ We would have $\displaystyle ab+bc+ca=-\frac{a^2+b^2+c^2}{2}\le 0,\,$ which together with the stipulation $ab+bc+ca\ge 0,\,$ implies $ab+bc+ca=0,\,$ so that $a^2+b^2+c^2\,$ and, subsequently, $a=b=c=0.$

Case 2: WLOG, $a+b+c=3,\,$ (in particular $a+b+c\gt 0).\,$ From $ab+bc+ca\ge 0,\,$ $ab+bc+ca=3(1-t^2),\,$ $0\ge t\ge 1.\,$ Also, from $abc\ge 0,\,$ we get

$\displaystyle \min (abc)=\begin{cases} (1+t)^2(1-2t), & if\,0\le t\le\frac{1}{2}\\ 0, & if \frac{1}{2}\le t\le 1. \end{cases}$

Using the formula,

$\small{\displaystyle\prod_{cycl}(a^2+ab+b^2)=(a+b+c)^2(ab+bc+ca)^2-(ab+bc+ca)^3-abc(a+b+c)^3,}$

we need to prove that $\displaystyle (1-t^2)^2(2+t^2)-abc\le\frac{81}{64}.\,$ If $0\le t\le\frac{1}{2},\,$ suffice it to show that

$\displaystyle (1-t^2)^2(2+t^2)-(1+t)^2(1-2t)\le\frac{81}{64},\,$ which is true since the left-hand side function $f(t)\,$ is increasing on $\displaystyle\left[0,\frac{1}{2}\right]\,$ and $\displaystyle f\left(\frac{1}{2}\right)=\frac{81}{64}.$

If $\displaystyle\frac{1}{2}\le t\le 1,\,$ then $(1-t^2)^2(2+t^2)-abc\le (1-t^2)^2(2+t^2)\,$ and, since the function $g(t)=(1-t^2)^2(2+t^2)\,$ is decreasing on $\displaystyle\left[\frac{1}{2},1\right],\,$ and $\displaystyle g\left(\frac{1}{2}\right)=\frac{81}{64},\,$ the proof is complete.

### Acknowledgment

The problem and the solution have been kindly communicated by Leo Giugiuc. The problem, due to Nguyen Van Huyen, has been posted at the artofproblemsolving forum, along with Leo's solution and a discussion that related this problem to another one of geometric nature. The appearance of the problem and the method of solution show a link to yet another problem.

- A Cyclic But Not Symmetric Inequality in Four Variables
- An Inequality with Constraint
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- 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
- 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

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