Israel tst 2011 geometrical inequality

Question

Inside an equilateral triangle of area $S$ lies a point, whose distances to the vertices are $x,y, z$. Prove that $xy + yz + zx \geq \frac{4}{\sqrt{3}} S$

I haven't got any idea yet. But I guess Fermat's point can be helpful....($x+y+z\ge \sqrt{3}a$)

Answer 1

Let our point be a origin of the Gauss plane, $\Delta ABC$ be our triangle,

$A(a)$, $B(b)$ and $C(c)$.
Since $\sum\limits_{cyc}\frac{ab}{(b-c)(c-a)}=\frac{\sum\limits_{cyc}(a^2b-a^2c)}{\prod\limits_{cyc}(a-b)}=-1$, we obtain:

$1=|\sum\limits_{cyc}\frac{ab}{(b-c)(c-a)}|\leq\sum\limits_{cyc}\frac{|a|\cdot|b|}{|b-c|\cdot|c-a|}=\frac{xy+xz+yz}{AB^2}$

Done!