Question

In any triangle, the minimum value of $\dfrac{{{r_1}{r_2}{r_3}}}{{{r^3}}}$ is equal to
A. $1$
B. $9$
C. $27$
D. None of these

Answer 1

Hint: To solve the question, First understand the concept of in-radii and ex-radii.
Inradius is a radius of an Incircle which is inside the triangle
Exradii is the radius of a Excircles, excircle of the triangle is a circle lying outside the triangle
Consider $r$ is the Inradius and ${r_1}$,${r_2}$ and ${r_3}$ are the Exradii.
Use the fact that the geometric mean is greater than the harmonic mean.
G.M. > H.M.
${\left( {\dfrac{{{r_1}}}{r} \cdot \dfrac{{{r_2}}}{r} \cdot \dfrac{{{r_3}}}{r}} \right)^{\dfrac{1}{n}}} > \dfrac{3}{{\dfrac{{{r_1}}}{r} + \dfrac{{{r_2}}}{r} + \dfrac{{{r_3}}}{r}}} \ldots (1)$
Since ${r_1} = \dfrac{\Delta }{{s - a}}$ , ${r_2} = \dfrac{\Delta }{{s - b}}$,${r_3} = \dfrac{\Delta }{{s - c}}$ and $r = \dfrac{\Delta }{s}$, substitute these values into the right side of inequality $(1)$ and solve the inequality.

Complete step-by-step solution:
Consider $r$ is the Inradius and ${r_1}$,${r_2}$ and ${r_3}$are the Exradii.
Inradius is a radius of a Incircle which is inside the triangle
Exradii is the radius of a Excircles, excircle of the triangle is a circle lying outside the triangle
Here, ${r_1} = \dfrac{\Delta }{{s - a}}$ , ${r_2} = \dfrac{\Delta }{{s - b}}$,${r_3} = \dfrac{\Delta }{{s - c}}$ and $r = \dfrac{\Delta }{s}$.
The Geometric mean: The geometric mean is defined as the nth root of the product of a set of numbers${x_1},{x_2},{x_3}, \ldots ,{x_n}$.
G.M.= ${\left( {\prod\limits_{i = 1}^n {{x_i}} } \right)^{\dfrac{1}{n}}}$
The Harmonic Mean: The harmonic mean formula is,
H.M.=$\dfrac{n}{{\sum\limits_{i = 1}^n {\dfrac{1}{{{x_i}}}} }}$
n=the number of the values in a dataset
${x_i}$=the point in a dataset
Use the fact that the geometric mean is greater than the harmonic mean.
G.M. $ \geqslant $ H.M.
Here, $n = 3$, according to the formula of G.M. and H.M.
\[{\left( {\dfrac{{{r_1}}}{r} \cdot \dfrac{{{r_2}}}{r} \cdot \dfrac{{{r_3}}}{r}} \right)^{\dfrac{1}{3}}} \geqslant \dfrac{3}{{\dfrac{r}{{{r_1}}} + \dfrac{r}{{{r_2}}} + \dfrac{r}{{{r_3}}}}} \ldots (1)\]
Substitute ${r_1} = \dfrac{\Delta }{{s - a}}$ , ${r_2} = \dfrac{\Delta }{{s - b}}$,${r_3} = \dfrac{\Delta }{{s - c}}$ and $r = \dfrac{\Delta }{s}$ values into inequality$(1)$
${\left( {\dfrac{{{r_1}}}{r} \cdot \dfrac{{{r_2}}}{r} \cdot \dfrac{{{r_3}}}{r}} \right)^{\dfrac{1}{3}}} \geqslant \dfrac{3}{{\dfrac{{\dfrac{\Delta }{s}}}{{\dfrac{\Delta }{{s - a}}}} + \dfrac{{\dfrac{\Delta }{s}}}{{\dfrac{\Delta }{{s - b}}}} + \dfrac{{\dfrac{\Delta }{s}}}{{\dfrac{\Delta }{{s - c}}}}}}$
\[{\left( {\dfrac{{{r_1}}}{r} \cdot \dfrac{{{r_2}}}{r} \cdot \dfrac{{{r_3}}}{r}} \right)^{\dfrac{1}{3}}} \geqslant \dfrac{3}{{\dfrac{{s - a}}{s} + \dfrac{{s - b}}{s} + \dfrac{{s - c}}{s}}}\]
Take L.C.M of the denominator of the right side of the inequality.
\[{\left( {\dfrac{{{r_1}}}{r} \cdot \dfrac{{{r_2}}}{r} \cdot \dfrac{{{r_3}}}{r}} \right)^{\dfrac{1}{3}}} \geqslant \dfrac{3}{{\dfrac{{s - a + s - b + s - c}}{s}}}\]
\[{\left( {\dfrac{{{r_1}}}{r} \cdot \dfrac{{{r_2}}}{r} \cdot \dfrac{{{r_3}}}{r}} \right)^{\dfrac{1}{3}}} \geqslant \dfrac{{3s}}{{s - a + s - b + s - c}}\]
\[{\left( {\dfrac{{{r_1}}}{r} \cdot \dfrac{{{r_2}}}{r} \cdot \dfrac{{{r_3}}}{r}} \right)^{\dfrac{1}{3}}} \geqslant \dfrac{{3s}}{{3s - (a + b + c)}} \ldots (2)\]
Here, s is the half of the perimeter of the triangle, $s = \dfrac{{a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c}}{2}$.
$ \Rightarrow 2s = a{\text{ + }}b{\text{ + }}c$
Substitute $2s = a{\text{ + }}b{\text{ + }}c$ into the equation$(2)$.
\[ \Rightarrow {\left( {\dfrac{{{r_1}}}{r} \cdot \dfrac{{{r_2}}}{r} \cdot \dfrac{{{r_3}}}{r}} \right)^{\dfrac{1}{3}}} \geqslant \dfrac{{3s}}{{3s - 2s}}\]
\[ \Rightarrow {\left( {\dfrac{{{r_1}}}{r} \cdot \dfrac{{{r_2}}}{r} \cdot \dfrac{{{r_3}}}{r}} \right)^{\dfrac{1}{3}}} \geqslant 3\]
\[ \Rightarrow {\left( {\dfrac{{{r_1}{r_2}r}}{{{r^3}}}} \right)^{\dfrac{1}{3}}} \geqslant 3\]
Take cube on both sides of the inequality,
\[ \Rightarrow {\left[ {{{\left( {\dfrac{{{r_1}{r_2}r}}{{{r^3}}}} \right)}^{\dfrac{1}{3}}}} \right]^3} \geqslant {\left( 3 \right)^3}\]
\[ \Rightarrow \dfrac{{{r_1}{r_2}r}}{{{r^3}}} \geqslant 27\]
$\therefore $ The minimum value of $\dfrac{{{r_1}{r_2}{r_3}}}{{{r^3}}}$ is $27$.

Option C is the correct answer.

Note: The important step is the note that the geometric mean is \[{\left( {\dfrac{{{r_1}}}{r} \cdot \dfrac{{{r_2}}}{r} \cdot \dfrac{{{r_3}}}{r}} \right)^{\dfrac{1}{3}}}\]not simply \[{\left( {{r_1} \cdot {r_2} \cdot {r_3}} \right)^{\dfrac{1}{3}}}\].
And ,the harmonic mean we use is \[\dfrac{r}{{{r_1}}} + \dfrac{r}{{{r_2}}} + \dfrac{r}{{{r_3}}}\].
There are three exradii so we take geometric and harmonic mean of three terms $\therefore n = 3$ .