Lets Play with Functions

We Should write it in another form i.e we can write a+b+c as f(1) and $\large\frac { f\left( -2 \right) -f\left( 1 \right) }{ 3 } =b-a$ so,

$\large\large\quad \frac { a+b+c }{ b-a } =\frac { 3f\left( 1 \right) }{ f\left( 1 \right) -f\left( -2 \right) } =\frac { 3 }{ 1-\large\frac { f(-2) }{ f\left( 1 \right) } }$

$\therefore for\quad given \quad expression\quad to \quad be \quad maximum \quad f\left( -2 \right) =0$

So, the minimum value of the expression is 3.