Minimum value#3

From $a,b\ge 0$ we have ${ \left( a-b \right) }^{ 2 }\left( a+b \right) \ge 0$

$\Longrightarrow { a }^{ 3 }+{ b }^{ 3 }\ge { a }^{ 2 }b+a{ b }^{ 2 }\quad \left( 1 \right)$

From $\left( 1 \right)$ we get ${ 3\left( { a }^{ 3 }+{ b }^{ 3 } \right) }\ge { 3\left( { a }^{ 2 }b+a{ b }^{ 2 } \right) }$ $\Longrightarrow { 4\left( { a }^{ 3 }+{ b }^{ 3 } \right) }\ge { { \left( a+b \right) }^{ 3 } }$

Similarly, ${ 4\left( { b }^{ 3 }+{ c }^{ 3 } \right) }\ge { { \left( b+c \right) }^{ 3 } }$ and ${ 4\left( { c }^{ 3 }+{ a }^{ 3 } \right) }\ge { { \left( c+a \right) }^{ 3 } }$

So, $C=\sqrt [ 3 ]{ 4\left( { a }^{ 3 }+{ b }^{ 3 } \right) } +\sqrt [ 3 ]{ 4\left( { b }^{ 3 }+{ c }^{ 3 } \right) } +\sqrt [ 3 ]{ 4\left( { c }^{ 3 }+{ a }^{ 3 } \right) } \ge \left( a+b \right) +\left( b+c \right) +\left( c+a \right) =2\times 2017=4034$

${ C }_{ min }=4034$ when $a=b=c=\frac { 2017 }{ 3 }$

Note: Inequality $\left( 1 \right)$ is the key.