# Dan Sitaru's Cyclic Inequality in Three Variables VIII

### Solution

Define $O=(0,0),$ $A=(x,0),$ $\displaystyle B=\left(-\frac{y}{2},\frac{y\sqrt{3}}{2}\right),$ $\displaystyle C=\left(-\frac{z}{2},-\frac{z\sqrt{3}}{2}\right).$ Then in $\Delta ABC,$ $a=BC=\sqrt{y^2+yz+z^2},$ $b=AC=\sqrt{z^2+zx+x^2},$ $c=AB=\sqrt{x^2+xy+y^2},$ as $O$ appears to be Fermat's point of $\Delta ABC.$ Further, if $S=[\Delta ABC],$

$4S=4(\Delta OAB]+[\Delta OBC]+[\Delta OCA])=\sqrt{3}(xy+yz+zx).$

Hence, the required inequality reduces to $\displaystyle \frac{8Ss}{\sqrt{3}}\le 3abc$ which is equivalent to $\displaystyle s\le\frac{3R\sqrt{3}}{2},$ which is well known. $(s$ here is the semiperimeter $\displaystyle s=\frac{a+b+c}{2}$ and $R$ the circumradius of $\Delta ABC.)$

### Acknowledgment

Leo Giugiuc has kindly posted problem by Dan Sitaru at the CutTheKnotMath facebook page, with a solution of his. The problem appeared earlier at the Romanian Mathematical Magazine.