Interesting Inequality

Because 'a' is the smallest number, for any arbitrary real number, ${ x }_{ i }-a\ge 0$ . Similarly, because 'b' is the largest number, for any arbitrary real number, ${ x }_{ i }-b\le 0$ . Multiplying those 2 inequalities gives us $({ x }_{ i }-a)({ x }_{ i }-b)={ { x }_{ i } }^{ 2 }-(a+b){ x }_{ i }+ab\le 0$ . Therefore, ${ { x }_{ 1 } }^{ 2 }-(a+b){ x }_{ 1 }+ab\le 0,\quad { { x }_{ 2 } }^{ 2 }-(a+b){ x }_{ 2 }+ab\le 0,...,\quad { { x }_{ n } }^{ 2 }-(a+b){ x }_{ n }+ab\le 0$ . Adding all of them gives $\sum _{ i=1 }^{ n }{ { { x }_{ i } }^{ 2 } } -(a+b)\sum _{ i=1 }^{ n }{ { x }_{ i } } +nab\le 0\rightarrow 1+nab\le 0\rightarrow ab\le -\frac { 1 }{ n }$