# An Acyclic Inequality in Three Variables

### Solution

\begin{align} &\left(\sum_{cycl}a^2-\sum_{cycl}ab\right)\left(\sum_{cycl}a^2+\sum_{cycl}ab\right)\\ &\qquad\qquad=\left(\sum_{cycl}a^2\right)^2-\left(\sum_{cycl}ab\right)^2\\ &\qquad\qquad=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2-\\ &\qquad\qquad\qquad\qquad-a^2b^2-a^2c^2-b^2c^2-2abc(a+b+c)\\ &\qquad\qquad=(a^4-2a^2bc+b^2c^2)+(b^4-2b^2ac+a^2c^2)\\ &\qquad\qquad\qquad\qquad+(c^4-2c^2ab+b^2a^2)\\ &\qquad\qquad=\sum_{cycl}(a^2-bc)^2 \end{align}

It follows that

\displaystyle \begin{align} &\frac{(a^2-bc)^2+(b^2-ca)^2+(c^2-ab)^2}{a^2+b^2+c^2+ab+bc+ca}\\ &\qquad\qquad=\frac{\displaystyle (\sum_{cycl}a^2-\sum_{cycl}ab)(\sum_{cycl}a^2+\sum_{cycl}ab)}{\displaystyle \sum_{cycl}a^2+\sum_{cycl}ab}\\ &\qquad\qquad=\sum_{cycl}a^2-\sum_{cycl}ab \end{align}

Suffice it to prove that

\displaystyle\begin{align} &\sum_{cycl}a^2-\sum_{cycl}ab\geq 3(a-b)(b-c)\,\Longleftrightarrow\\ &a^2+b^2+c^2-ab-bc-ca\geq 3ab-3ac-3b^2+3bc\,\Longleftrightarrow\\ &a^2+4b^2+c^2-4ab-4bc+2ca\geq 0\,\Longleftrightarrow\\ &(a-2b+c)^2\geq 0 \end{align}

Equality is attained for $\displaystyle b=\frac{a+c}{2}.$

### Remark

It may be worth noting that he inequality at hand is not cyclic, and, in this sense, the appearance of the cyclic sums may lead the reader to think otherwise. The fact is that the three variables do not occur in the inequality in a symmetric manner, so that an argument that relies on the "WLOG" reasoning may not be valid. For example, if we assume - apparently WLOG - that $a\ge c\ge b,\,$ then the proof becomes immediate because

$\displaystyle \frac{(a^2-bc)^2+(b^2-ca)^2+(c^2-ab)^2}{a^2+b^2+c^2+ab+bc+ca}\geq 0\ge 3(a-b)(b-c).$

However, such a proof is faulty.

### Acknowledgment

The problem above has been posted on the CutTheKnotMath facebook page and the solution above communicated to me by Dan Sitaru. Originally, the problem has been published at the Romanian Mathematical Magazine.