# Estimate for $\frac{1}{\Delta}\sqrt{abc(a+b+c)}$ [duplicate]

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....

• Note that the numerator is equal to the denominator for an equilateral triangle, and the denominator gets really small as the triangle degenerates to a line. That should tell you that the answer is $\geq4$. Now for proving it... Aug 9 '15 at 8:54
• this shows only $\geq 4$ or $\le 4$ Aug 9 '15 at 8:56
• the magic word is here RAVI-substitution Aug 9 '15 at 8:58
• @Dr.SonnhardGraubner,Sir what is RAVI-substitution? Aug 9 '15 at 8:59

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 $$\sqrt{2} \sqrt{\frac{(x+y) (x+z) (y+z)}{x y z}}\geq \sqrt{8}\sqrt{2}=4$$