Question

If a $△ABC$ has altitudes $h_1,h_2,h_3$ and inradius $r$, then the minimum value of $\frac{h_1+r}{h_1-r}+\frac{h_2+r}{h_2-r}+\frac{h_3+r}{h_3-r}$ is

Answer 1

We have,

$△=\frac{1}{2}{ah}_1\Rightarrow h_1=\frac{2△}{a}$

Similarly, $h_2=\frac{2△}{b}$ and

$h_3=\frac{2△}{c}$

$\therefore \frac{h_1+r}{h_1-r}+\frac{h_2+r}{h_2-r}+\frac{h_3+r}{h_3-r}$

$=\frac{2\frac{△}{a}+\frac{△}{s}}{2\frac{△}{a}-\frac{△}{s}}+\frac{2\frac{△}{b}+\frac{△}{s}}{2\frac{△}{b}-\frac{△}{s}}+\frac{2\frac{△}{c}+\frac{△}{s}}{2\frac{△}{c}-\frac{△}{s}}$

$=\frac{2s+a}{2s-a}+\frac{2s+b}{2s-b}+\frac{2s+c}{2s-c}$

$=(\frac{2s+a}{2s-a}+1)+(\frac{2s+b}{2s-b}+1)+(\frac{2s+c}{2s-c}+1)-3$

$=\frac{4s}{2s-a}+\frac{4s}{2s-b}+\frac{4s}{2s-c}-3$

$=3\left[\frac{1}{3}\{\frac{4s}{2s-a}+\frac{4s}{2s-b}+\frac{4s}{2s-c}\}\right]-3$

$\ge 3\left[\frac{3}{\frac{2s-a}{4s}+\frac{2s-b}{4s}+\frac{2s-c}{4s}}\right]-3(\because A.M.\ge H.M.)$

$=3\left[\frac{3}{\frac{6s-2s}{4s}}\right]-3=9-3=6$

$\therefore$ Min. value $=6$