RMO 2014

Nihar Mahajan , please edit the question to

$\large{K\sum _{ j=1 }^{ 2014 }{ \frac { { x }_{ j }^{ 2 } }{ 1-{ x }_{ j } } } \ge 1}$

By Cauchy-Schwarz Inequality/Titu's lemma, we have

$\large{\sum _{ j=1 }^{ 2014 }{ \frac { { x }_{ j }^{ 2 } }{ 1-{ x }_{ j } } } \ge \frac { { \left( \sum _{ j=1 }^{ 2014 }{ { x }_{ j } } \right) }^{ 2 } }{ \sum _{ j=1 }^{ 2014 }{ \left( 1-{ x }_{ j } \right) } } \ge \frac { 1 }{ 2014-\sum _{ j=1 }^{ 2014 }{ { x }_{ j } } } \ge \frac { 1 }{ 2013 } \\2013\sum _{ j=1 }^{ 2014 }{ \frac { { x }_{ j }^{ 2 } }{ 1-{ x }_{ j } } \ge } 1}$

this gives the smallest value of constant $2013$ .

Also can be done by M.S-A.M-G.M-H.M