How to prove : If $a,b,c$ are in H.P. in $\Delta ABC$ then $\frac{(\sum ab-2bc ) ( \sum ab-2ca)}{(abc)^2} = \frac{\sin A}{\Delta} -\frac{1}{b^2} $

Question

How to prove : If $a,b,c$ are in H.P. in $\Delta ABC$ then $$\frac{(\sum ab-2bc ) ( \sum ab-2ca)}{(abc)^2}=\frac{\sin A}{\Delta} -\frac{1}{b^2} $$

Please guide how to proceed , will be of great help. Thanks.

Answer 1

As $$\triangle=\dfrac{bc\sin A}2$$

So,we need $$\dfrac{(ab+ca-bc)(ab+bc-ca)}{(abc)^2}=\dfrac{2b-c}{b^2c}$$

$$\iff(ab)^2-(bc-ca)^2=a^2c(2b-c)$$

$$\iff a^2b^2-b^2c^2=2a^2bc-2abc^2$$

$$\iff b(a^2-c^2)=2ca(a-c)$$

As in general, $a-c\ne0$ unless $a=b=c$ $$b(a+c)=2ca\iff\dfrac1a+\dfrac1c=\dfrac2b$$