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

Question

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... — Arthur, Aug 09 '15 at 08:54

score 3 · Accepted Answer · answered Aug 09 '15 at 08:55

3

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

answered Aug 09 '15 at 08:55

Dr. Sonnhard Graubner

95,283

see here http://math.stackexchange.com/questions/1174962/ravi-substitution-in-inequalities – Dr. Sonnhard Graubner Aug 09 '15 at 09:01

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

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