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) 



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*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

The actual inequality is $(abc)^2\ge \left(\dfrac{4\Delta}{\sqrt{3}}\right)^3$.I will show this one.


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*Lemma:
$\displaystyle \sin{\alpha}+\sin{\beta}+\sin{\gamma} \le \frac{3\sqrt{3}}{2}$
when $\alpha$,$\beta$,$\gamma$ are angles of a triangle.Now we see
that $\sin x$ is concave in $(0,\pi)$ so applying Jensen's inequality
we get $\dfrac{\sin{\alpha}+\sin{\beta}+\sin{\gamma}}{3}\le
   \sin\left({\dfrac{\alpha+\beta+\gamma}{3}}\right)=\dfrac{\sqrt{3}}{2}$
So our lemma is proved.
Now to the actual problem.$a+b+c=2R(\sin{\alpha}+\sin{\beta}+\sin{\gamma})\le 3\sqrt{3}R$ (from the lemma).
Now $$\dfrac{4\Delta}{\sqrt{3}}=\dfrac{abc}{R\sqrt{3}}\le \dfrac{3abc}{a+b+c}\le (abc)^{2/3}$$ where in the last step we have used AM-GM.Now cubing both sides we get the desired inequality.
