$\eqalign{
& co\;{\left( {x + y} \right)^2} \ge 0 \to {x^2} + {y^2} \ge - 2xy \cr
& \to 1 = {x^2} + {y^2} - xy \ge - 2xy - xy = - 3xy \to xy \ge - {1 \over 3}\;\left( {dau = \leftrightarrow x = - y\;tu\;tinh\;nhe} \right) \cr
& co\;{\left( {x - y} \right)^2} \ge 0 \to {x^2} + {y^2} \ge 2xy \cr
& \to 1 = {x^2} + {y^2} - xy \ge 2xy - xy = xy \to xy \le 1\;\left( {dau = \leftrightarrow x = y\;tu\;tinh} \right) \cr
& co\;{x^2} + {y^2} = 1 + xy \ge {2 \over 3}\;\left( {do\;xy \ge - {1 \over 3}} \right) \cr
& \cos i: \cr
& {{{x^4} + {y^4}} \over 2} \ge {x^2}{y^2} \cr
& {x^4} + {1 \over 9} \ge {{2{x^2}} \over 3}\;\left( 1 \right) \cr
& {y^4} + {1 \over 9} \ge {{2{y^2}} \over 3}\;\left( 2 \right) \cr
& \left( 1 \right)\left( 2 \right) \to {x^4} + {y^4} \ge {{2\left( {{x^2} + {y^2}} \right)} \over 3} - {2 \over 9} \ge {4 \over 9} - {2 \over 9} = {2 \over 9} \cr
& \to {x^4} + {y^4} - {x^2}{y^2} = {{{x^4} + {y^4}} \over 2} - {x^2}{y^2} + {{{x^4} + {y^4}} \over 2} \ge 0 + {1 \over 2}*{2 \over 9} = {1 \over 9} \cr
& dau = \leftrightarrow .... \cr
& \cr} $
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn