Minimum value#3

Let $BC=a,CA=b,AB=c$ , then we have $a,b,c>0 , { a }^{ 2 }={ b }^{ 2 }+{ c }^{ 2 }\ge \frac { { \left( b+c \right) }^{ 2 } }{ 2 }$ (by applying Pythagorean theorem and AM-GM inequality).

$\Longrightarrow \frac { b+c }{ a } \le \sqrt { 2 } \quad \left( 1 \right)$ .

By applying Angle bisector theorem,

$\frac { AF }{ BF } =\frac { b }{ a } \Longrightarrow \frac { AF }{ AB } =\frac { b }{ a+b } =\frac { AF }{ c } \Longrightarrow AF=\frac { bc }{ a+b }$ .

Similarly, $AE=\frac { bc }{ a+c } .$

Also, we notice that

${ Area }_{ ABC }={ Area }_{ ABD }+{ Area }_{ ADC }\\ \Longrightarrow bc=AD\left( c\sin { \widehat { ABD } +b\sin { \widehat { CAD } } } \right) =AD\times \frac { 1 }{ \sqrt { 2 } } \left( b+c \right) \\ \Longrightarrow AD=\frac { \sqrt { 2 } bc }{ b+c }$ .

By similarly, $AK=\frac { \sqrt { 2 } AE.AF }{ AE+AF } =\frac { \sqrt { 2 } bc }{ 2a+b+c } .$

So,

$\frac { AK }{ AD } =\frac { b+c }{ 2a+b+c } \Longrightarrow \frac { MN }{ a } =\frac { b+c }{ 2a+b+c } \\ \Longrightarrow MN=\frac { a\left( b+c \right) }{ 2a+b+c } =\frac { b+c }{ 2+\frac { b+c }{ a } } \ge \frac { b+c }{ 2+\sqrt { 2 } }$ (from $(1)$ )

$\Longrightarrow MN\ge \frac { 2-\sqrt { 2 } }{ 2 } \left( AB+AC \right) \Longrightarrow \frac { MN }{ AB+AC } \ge \frac { 2-\sqrt { 2 } }{ 2 } =1-\frac { \sqrt { 2 } }{ 2 }$

Hence, ${ \left( \frac { MN }{ AB+AC } \right) }_{ min }=1-\frac { \sqrt { 2 } }{ 2 }$ when $b=c=\frac { a\sqrt { 2 } }{ 2 }$