Maximum value in triangle#1

There are 2 formulas related to $r$ and $R$ should be using:

+) The radius of triangle's incircle is given by

$r=\frac { 2\times Area }{ Perimeter } =\frac { 2\sqrt { p\left( p-a \right) \left( p-b \right) \left( p-c \right) } }{ a+b+c }$ (applying Heron's Formula for the area of a triangle where $p$ is half the perimeter, or $\frac { a+b+c }{ 2 }$ )

$\Longrightarrow r=\frac { 1 }{ 2 } \sqrt { \frac { \left( b+c-a \right) \left( a+c-b \right) \left( a+b-c \right) }{ a+b+c } }$

+) The radius of triangle's circumcircle is given by

$R=\frac { abc }{ \sqrt { \left( a+b+c \right) \left( b+c-a \right) \left( a+c-b \right) \left( a+b-c \right) } }$

From 2 above formulas, $\frac { r }{ R } =\frac { \left( b+c-a \right) \left( a+c-b \right) \left( a+b-c \right) }{ abc }$

Let $f\left( c \right) =\frac { \left( b+c-a \right) \left( a+c-b \right) \left( a+b-c \right) }{ abc }$

$\Longrightarrow f^{ ' }\left( c \right) =\frac { \left( a+b \right) { c }^{ 2 }-2{ c }^{ 3 }+\left( a+b \right) { \left( a-b \right) }^{ 2 } }{ 2ab{ c }^{ 2 } } \ge \frac { \left( a+b-2c \right) { c }^{ 2 } }{ 2ab{ c }^{ 2 } } > 0$

So, $f\left( c \right)$ is increasing in term of $c$ .

Since $a+b\ge 3c\Longrightarrow c\le \frac { a+b }{ 3 }$

$\Longrightarrow f\left( c \right) \le f\left( \frac { a+b }{ 3 } \right) =\frac { 4\left( 2b-a \right) { \left( 2a-b \right) } }{ 9ab } =\frac { 4 }{ 9 } -\frac { 2{ \left( a-b \right) }^{ 2 } }{ 9ab } \le \frac { 4 }{ 9 }$ , or $\frac { r }{ R } \le \frac { 4 }{ 9 }$

We then have ${ \left( \frac { r }{ R } \right) }_{ max }=\frac { 4 }{ 9 }$ when $a=b=\frac { 3 }{ 2 } c$ .