# Inequality with Three Numbers, Not All Zero

### Proof

We have $2(a-b)^2\ge 0,\;$ implying

$3(a^2-ab+b^2)\ge a^2+ab+b^2.$

Similarly,

$3(c^2-ca+a^2)\ge c^2+ca+a^2\;$ and
$3(b^2-bc+c^2)\ge b^2+bc+c^2.$

It follows (via the AM-GM inequality) that

\displaystyle\begin{align} 3\sum_{cycl}\frac{a^2-ab+b^2}{b^2+bc+c^2} &\ge \sum_{cycl}\frac{a^2+ab+b^2}{b^2+bc+c^2}\\ &\ge 3\sqrt[3]{\prod_{cycl}\frac{a^2+ab+b^2}{b^2+bc+c^2}}\\ &= 3. \end{align}

This completes the proof.

### An Illustration

Graphical view of the problem: $\gt 1,\;$ except in a tiny nugget around $\{0,0,0\}.$

### Extra

Let $a,b,c\;$ be the side lengths of $\Delta ABC.\;$ Then

$\displaystyle\sum_{cycl}\frac{a^2-ab+b^2}{b^2+bc+c^2}\lt 3.$

Indeed, each of the three fractions in the sum is less than $1.\;$ For example, $\displaystyle\frac{a^2-ab+b^2}{b^2+bc+c^2}\lt 1\;$ is equivalent to $a^2-ab+b^2\lt b^2+bc+c^2,\;$ which, in turn, is equivalent to the obvious $(a+c)(a-b-c)\lt 0.$

### Acknowledgment

The problem (by Nguyen Viet Hung) was communicated to me by Leo Giugiuc, along with his solution. The illustration above is by Nassim N. Taleb.