Mathematics - Properties of Triangle Question with Solution | TestHub

$\frac{h_1+r}{h_1-r}+\frac{h_2+r}{h_2-r}+\frac{h_3+r}{h_3-r}\ge 6$ $\Delta =\frac{1}{2}\mathrm{ah}\Rightarrow {\mathrm{h}}_1=\frac{2\Delta }{\mathrm{a}}$ Similarly ${\mathrm{h}}_1=\frac{2\Delta }{\mathrm{b}},{\mathrm{h}}_3=\frac{2\Delta }{\mathrm{c}}$ So $\frac{{\mathrm{h}}_1+\mathrm{r}}{{\mathrm{h}}_1-\mathrm{r}}+\frac{{\mathrm{h}}_2+\mathrm{r}}{{\mathrm{h}}_2-\mathrm{r}}=\frac{{\mathrm{h}}_3+\mathrm{r}}{{\mathrm{h}}_3-\mathrm{r}}=\frac{2\Delta /\mathrm{a}+\Delta /\mathrm{s}}{2\Delta /\mathrm{a}-\Delta /\mathrm{s}}+\frac{2\Delta /\mathrm{b}+\Delta /\mathrm{s}}{2\Delta /\mathrm{c}-\Delta /\mathrm{s}}+\frac{2\Delta /\mathrm{c}+\Delta /\mathrm{s}}{2\Delta /\mathrm{c}-\Delta /\mathrm{s}}$ $=\frac{2s+a}{2s-a}=\frac{2s+b}{2s-b}+\frac{2s+c}{2s-c}=\frac{4s}{2s-a}+\frac{4s}{2s-b}+\frac{4s}{2s-c}-3$ $=3\left[\frac{1}{3}\left\{\frac{4s}{2s-a}+\frac{4s}{2s-b}+\frac{4s}{2s-c}\right\}\right]-3\ge 3\left(\frac{3}{\frac{2s-a}{4s}+\frac{2s-b}{4s}+\frac{2s-c}{4s}}\right]=3$ Since $(AM\ge HM)\ge 6$