21st Mediterranean Mathematical Competition

It is well known that $AD=\frac{2bc}{b+c}\cos \frac{A}{2}=\frac{bc\sqrt{3}}{b+c}$. Let $h=AM$ be the length of the altitude from $A$ in the triangle $ABC$. From the theorem of the internal bisector we get $$BD=\frac{ac}{b+c},CD=\frac{ab}{b+c}.$$ Let $p_{ABD}=\frac{BD+DA+AB}{2}$ be the half perimeter of triangle $ABD$. If we denote $[ABD]$ the surface of the triangle $ABD$, then $$r_B=\frac{[ABD]}{p_{ABD}}=\frac{h\cdot BD}{2\cdot p_{ABD}}=\frac{h\cdot BD}{BD+AD+AB}=\frac{\frac{h\cdot ac}{b+c}}{\frac{ac}{b+c}+\frac{bc\sqrt{3}}{b+c}+c}=\frac{ah}{2p+\sqrt{3}\cdot b}.$$ Analogously we get $r_C=\frac{ah}{2p+\sqrt{3}\cdot c}$, where $p=\frac{a+b+c}{2}$. From this we have $$\frac{1}{r_B}+\frac{1}{r_C}=\frac{2p+b\sqrt{3}}{ah}+\frac{2p+c\sqrt{3}}{ah}=\frac{4p+(b+c)\sqrt{3}}{ah}.$$ But $r=\frac{[ABC]}{p}=\frac{ah}{2p}$, so $ah=2pr$. Thus, $$\frac{1}{r_B}+\frac{1}{r_C}=\frac{4p+(b+c)\sqrt{3}}{2pr}.$$ Now, $b=AC$, $c=AB$, and $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$, so $[ABC]=\frac{1}{2}bc\sin 60^{\circ }=\frac{bc\sqrt{3}}{4}$, and $r=\frac{[ABC]}{p}=\frac{bc\sqrt{3}}{4p}$, so $\frac{1}{r}=\frac{4p}{bc\sqrt{3}}$.

Also, $$\frac{1}{b}+\frac{1}{c}=\frac{b+c}{bc}.$$ Therefore, $$2\left(\frac{1}{r}+\frac{1}{b}+\frac{1}{c}\right)=2\left(\frac{4p}{bc\sqrt{3}}+\frac{b+c}{bc}\right)=\frac{8p}{bc\sqrt{3}}+\frac{2(b+c)}{bc}.$$ But $ah=2pr$, so $ah=2p\cdot \frac{bc\sqrt{3}}{4p}=\frac{bc\sqrt{3}}{2}$.

Now, substitute $ah$ into the earlier formula: $$\frac{1}{r_B}+\frac{1}{r_C}=\frac{4p+(b+c)\sqrt{3}}{ah}=\frac{4p+(b+c)\sqrt{3}}{\frac{bc\sqrt{3}}{2}}=\frac{2[4p+(b+c)\sqrt{3}]}{bc\sqrt{3}}=\frac{8p}{bc\sqrt{3}}+\frac{2(b+c)}{bc}.$$ This matches the previous expression, so $$\frac{1}{r_B}+\frac{1}{r_C}=2\left(\frac{1}{r}+\frac{1}{b}+\frac{1}{c}\right).$$