NMTC final test 2014

Setting $x=a+1$ , $y=b+1$ , $z=c+1$ forces the L.H.S. of the given inequality equivalent to

$\large \frac { { (a+1) }^{ 4 } }{ { b }^{ 2 } } +\frac { { (b+1) }^{ 4 } }{ { c }^{ 2 } } +\frac { { (c+1) }^{ 4 } }{ { a }^{ 2 } }$

and this is (by AM-GM )

$\large\ \ge 3\sqrt [ 3 ]{ \left\{ \frac { { (a+1) }^{ 4 }{ (b+1) }^{ 4 }{ (c+1) }^{ 4 } }{ { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 } } \right\} }$

Since for any positive integer $x$ , $x+1\ge 2\sqrt { x }$ , we have

$\large\ 3\sqrt [ 3 ]{ \left\{ \frac { { (a+1) }^{ 4 }{ (b+1) }^{ 4 }{ (c+1) }^{ 4 } }{ { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 } } \right\} } \ge \quad 3\sqrt [ 3 ]{ \frac { { (2\sqrt { a } ) }^{ 4 }2{ \sqrt { b } }^{ 4 }2{ \sqrt { c } }^{ 4 } }{ { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 } } } =48$

Where $K = 48.$

Hence answer: $\boxed{48}$