# Dorin Marghidanu's Cyclic Inequality in Three Variables

### Solution 1

Using complex numbers, $\sqrt{a^2-ab+b^2}=|a+\epsilon b|,\,$ $\sqrt{b^2-bc+c^2}=|b+\epsilon c|,\,$ $\sqrt{c^2-ca+a^2}=|c+\epsilon a|,\,$ where $\displaystyle\epsilon=\frac{-1+\sqrt{3}}{2}.\,$

With Minkowsky's Inequality, we obtain

\displaystyle\begin{align} \sum_{cycl}\sqrt{a^2-ab+b^2}\sqrt{b^2-bc+c^2}&=\sum_{cycl}|a+\epsilon b|\cdot |b+\epsilon c|\\ &\ge \left|\sum_{cycl}(a+\epsilon b)(b+\epsilon c)\right|\\ &= x^2+y^2+z^2. \end{align}

### Solution 2

Employing Cauchy's Inequality,

\displaystyle\begin{align} &\sqrt{a^2-ab+b^2}\sqrt{a^2-ac+c^2}\\ &\qquad\qquad=\small{\sqrt{\left[ \left(a-\frac{b}{2}\right)^2 +\left(\frac{b\sqrt{3}}{2}\right)^2\right]\left[\left(a-\frac{c}{2}\right)^2+\left(\frac{c\sqrt{3}}{2}\right)^2\right]}}\\ &\qquad\qquad\ge\left|\left(a-\frac{b}{2}\right)\left(a-\frac{c}{2}\right)+\frac{3bc}{4}\right|\\ &\qquad\qquad\ge \left(a-\frac{b}{2}\right)\left(a-\frac{c}{2}\right)+\frac{3bc}{4}\\ &\qquad\qquad=a^2-\frac{a(b+c)}{2}+bc. \end{align}

Two similar inequalities are derived in the same manner. Adding up, we get

\displaystyle\begin{align}\sum_{cycl}\sqrt{a^2-ab+b^2}\sqrt{b^2-bc+c^2}&\ge\sum_{cycl}\left(a^2-\frac{a(b+c)}{2}+bc\right)\\ &=a^2+b^2+c^2. \end{align}

### Solution 3

We have

\displaystyle\begin{align} &a^2-ab+b^2=\left(a-\frac{b}{2}\right)^2+\frac{3b^2}{4}=\left(b-\frac{a}{2}\right)^2+\frac{3a^2}{4},\\ &b^2-bc+c^2=\left(b-\frac{c}{2}\right)^2+\frac{3c^2}{4}=\left(c-\frac{b}{2}\right)^2+\frac{3b^2}{4},\\ &c^2-ca+a^2=\left(c-\frac{a}{2}\right)^2+\frac{3a^2}{4}=\left(a-\frac{c}{2}\right)^2+\frac{3c^2}{4}. \end{align}

With these relations, we get via the Cauchy-Schwarz inequality,

\displaystyle\begin{align} \sqrt{(a^2-ab+b^2)(b^2-bc+c^2)}&=\small{\sqrt{\left(a-\frac{b}{2}\right)^2+\frac{3b^2}{4}}\sqrt{\left(c-\frac{b}{2}\right)^2+\frac{3b^2}{4}}}\\ &\ge\left(a-\frac{b}{2}\right)\left(c-\frac{b}{2}\right)+\frac{3b^2}{4}. \end{align}

Equality is obtained if $\displaystyle \left(a-\frac{b}{2}\right)=\left(c-\frac{b}{2}\right)\,$ i.e., when $a=c,\,$ or when $b=0.\,$ Hence, we have

$\displaystyle\sqrt{(a^2-ab+b^2)(b^2-bc+c^2)}\ge b^2+ac-\frac{ab}{2}-\frac{bc}{2}$

and, similarly

\displaystyle\begin{align} \sqrt{(b^2-bc+c^2)(c^2-ca+a^2)}\ge c^2+ab-\frac{ac}{2}-\frac{bc}{2},\\ \sqrt{(c^2-ca+a^2)(a^2-ab+b^2)}\ge a^2+bc-\frac{ab}{2}-\frac{ac}{2}. \end{align}

Adding all three we prove the required inequality. Equality is attained for the triplets in the form $(k,k,k),\,$ $(k,0,0),\,$ $(0,k,0),\,$ $(0,0,k).\,$

### Acknowledgment

This problem has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu. Leo Giugiuc gave two solutions (Solutions 1 & 2). He also pointed to a similar problem of this form from 2012: $\displaystyle\sum_{cycl}\sqrt{a^2+ab+b^2}\sqrt{b^2+bc+c^2}\ge (a+b+c)^2.$ Solution 3 is by Dorin Marghidanu.