Cool Inequality #9 - 2 days to 2016, complex one

We have

$\Large{a\ge b\ge c\\ a+c\ge b+c\ge 2c\\ \\ \Longrightarrow a+c\ge b+c\\ \frac { 1 }{ b+c } \ge \frac { 1 }{ a+c } \\ \\ \Longrightarrow a+c\ge 2c\\ \frac { 1 }{ 2c } \ge \frac { 1 }{ a+c } \\ \\ \Longrightarrow b+c\ge 2c\\ \frac { 1 }{ b+c } \ge \frac { 1 }{ 2c } }$

From these have two sets

$\Large{a\ge b\ge c \\ \frac { 1 }{ 2c } \ge \frac { 1 }{ b+c } \ge \frac { 1 }{ c+a } }$

Using Rearrangement Inequality,

$\Large{\frac { a }{ 2c } +\frac { b }{ b+c } +\frac { c }{ c+a } \ge \frac { a }{ b+c } +\frac { b }{ c+a } +\frac { c }{ 2c } \\ \frac { 17 }{ 2 } -\frac { 1 }{ 2 } \ge \frac { a }{ b+c } +\frac { b }{ c+a } \\ \\ 8\ge \frac { a }{ b+c } +\frac { b }{ c+a } }$

Hence the maximum value is $\boxed{8}$