Proving that $(abc)^2\geq\left(\frac{4\Delta}{\sqrt{3}}\right)^3$, where $a$, $b$, $c$ are the sides, and $\Delta$ the area, of a triangle Let $a$, $b$, $c$ be the sides of $\triangle ABC$.

Prove  $$(abc)^2\geq\frac{4\Delta}{\sqrt{3}}$$ 
  where $\Delta$ is the area of the triangle. 
(Editor's note: As observed in the answers, the target relation is in error. It should be $$(abc)^2\geq\left(\frac{4\Delta}{\sqrt{3}}\right)^3$$ I incorporated the correction into the title. —@Blue) 



*

*Clearly, $a$, $b$, $c$ are the sides opposite to the angles $A$, $B$, $C$

*I considered the point $F$ inside the triangle as the Fermat point of the triangle and named $FA=x$, $FB=y$, and $FC=z$.

*Then $\angle AFB = \angle AFC = \angle BFC=120^\circ$, so we have $$a^2=y^2+z^2+yz$$ and similarly for $b^2$ and $c^2$.

*Observe that
$$xy+yz+zx=\frac{4\Delta}{\sqrt{3}}$$

*So, we have a big inequality to prove:
$$\left(x^2+xy+y^2\right)\left(y^2+yz+z^2\right)\left(z^2+zx+x^2\right)\geq xy+yz+zx$$ but I can't prove it.


Thanks for help

I think you mean to the following:
$$(abc)^{\frac{2}{3}}\geq\frac{4\Delta}{\sqrt3},$$ which is
$$16\Delta^2\leq3\sqrt[3]{a^4b^4c^4}$$ or
$$\sum_{cyc}(2a^2b^2-a^4)\leq\sqrt[3]{a^4b^4c^4}$$ or
$$\sum_{cyc}\left(a^4-2a^2b^2+\sqrt[3]{a^4b^4c^4}\right)\geq0,$$ which is true by Schur and AM-GM:
$$\sum_{cyc}\left(a^4-2a^2b^2+\sqrt[3]{a^4b^4c^4}\right)=\sum_{cyc}\left(\sqrt[3]{a^{12}}+\sqrt[3]{a^4b^4c^4}-2a^2b^2\right)\geq$$
$$\geq\sum_{cyc}\left(\sqrt[3]{a^9b^3}+\sqrt[3]{a^9c^3}-2a^2b^2\right)=\sum_{cyc}(a^3b+ab^3-2a^2b^2)\geq0.$$
