Estimate for $\frac{1}{\Delta}\sqrt{abc(a+b+c)}$ If the triangle ABC has sides $a,b,c$ opposite to the vertices A,B,C respectively and $\Delta$ is the area.The expression $\frac{1}{\Delta}\sqrt{abc(a+b+c)}$ is 
$(A)\leq16\hspace{1 cm}(B)\geq16\hspace{1 cm}(C)\leq4\hspace{1 cm}(D)\geq4$
Using Herons formula, $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$ and $s=\frac{a+b+c}{2}$, given expression reduces to
$4\sqrt{\frac{abc}{(a+b-c)(b+c-a)(a+c-b)}}$ and I could not solve further to find the answer. Please help....

with $a=y+z,b=x+z,c=x+y$ we get
$$\sqrt{2} \sqrt{\frac{(x+y) (x+z) (y+z)}{x y z}}$$
by AM-GM we have $$(x+y)(x+z)(y+z)\geq 8xyz$$
thus our term above is $$\sqrt{2} \sqrt{\frac{(x+y) (x+z) (y+z)}{x y z}}\geq \sqrt{8}\sqrt{2}=4$$
