Question

Prove that in a triangle $\frac{bc}{r_1}+\frac{ca}{r_2}+\frac{ab}{r_3}=2R[(\frac{a}{b}+\frac{b}{a})+(\frac{b}{c}+\frac{c}{b})+(\frac{c}{a}+\frac{a}{c})-3]$

Answer 1

$\frac{bc}{r_1}$+$\frac{ca}{r_2}$+$\frac{ab}{r_3}$

=$\frac{bc(s-a)}{△}$+$\frac{ca(s-b)}{△}$+$\frac{ab(s-c)}{△}$

=$\frac{bc(b+c-a)+ca(a-b+c)+ab(a+b-c)}{△}$

=$\left[\frac{2(a^2+b^2+c^2)-3\left(abc\right)}{\left(abc\right)}\right]\times 2R$

=$2R[(\frac{a}{b}+\frac{b}{a})+(\frac{b}{c}+\frac{c}{a})+(\frac{c}{a}+\frac{a}{c})-3]$