# An Inequality with Constraint XIV

### 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.