2014 Mania!

$Actually\quad the\quad solution\quad is\quad fairly\quad simple,\\ once\quad you\quad factor\quad out\quad \sqrt { 2014^{ 2013 } } from\quad the\quad numerator\quad and\quad denominator\quad you\quad get\quad the\quad following:\\ x\quad =\quad \frac { \sqrt { 2016 } -\sqrt { 2015 } }{ \sqrt { 2014 } -\sqrt { 2013 } } \quad y\quad =\quad \frac { \sqrt { 2016 } -\sqrt { 2015 } }{ \sqrt { 2015 } -\sqrt { 2014 } } \quad z\quad =\quad \frac { \sqrt { 2014 } -\sqrt { 2013 } }{ \sqrt { 2016 } -\sqrt { 2015 } } \\ Now\quad there\quad are\quad two\quad ways\quad to\quad solve\quad this,\quad one\quad is\quad to\quad plug\quad in\quad the\quad values,\quad another\quad is\quad to\quad realize\quad \\ that\quad as\quad the\quad numbers\quad get\quad bigger\quad they\quad get\quad less\quad further\quad apart.\quad For\quad example:\quad \sqrt { 2 } -\sqrt { 1 } >\sqrt { 3 } -\sqrt { 2 } .\\ With\quad that\quad in\quad mind,\quad \quad \sqrt { 2016 } -\sqrt { 2015 } \quad <\quad \sqrt { 2015 } -\sqrt { 2014 } \quad <\quad \sqrt { 2014 } -\sqrt { 2013 } \\ So,\quad \frac { \sqrt { 2016 } -\sqrt { 2015 } }{ \sqrt { 2014 } -\sqrt { 2013 } } <\frac { \sqrt { 2016 } -\sqrt { 2015 } }{ \sqrt { 2015 } -\sqrt { 2014 } } \quad becase\quad as\quad denominator\quad becomes\quad smaller\quad the\quad number\\ becomes\quad bigger\quad and\quad \frac { \sqrt { 2014 } -\sqrt { 2013 } }{ \sqrt { 2016 } -\sqrt { 2015 } } \quad is\quad the\quad only\quad number\quad greater\quad than\quad 1.\quad \\ So,\quad x\quad <\quad y\quad <\quad z$