bài 2: hơi dài tý
$\eqalign{
& dat\;a = 1 - x;\;b = 1 - y;\;c = 1 - z \cr
& \to - 1 \le a,b,c \le 1 \cr
& \to a + b + c = 0 \cr
& \to M = {\left( {a - 1} \right)^4} + {\left( {b - 1} \right)^4} + {\left( {c - 1} \right)^4} + 12abc \cr
& = \left( {{a^4} + {b^4} + {c^4}} \right) - 4\left( {{a^3} + {b^3} + {c^3}} \right) + 6\left( {{a^2} + {b^2} + {c^2}} \right) + 12abc + 3 \cr
& co:{a^4} + {b^4} + {c^4} = {\left( {{a^2} + {b^2} + {c^2}} \right)^2} - 2{a^2}{b^2} - 2{b^2}{c^2} - 2{c^2}{a^2}\;\left( 1 \right) \cr
& = {\left( {{{\left( {a + b + c} \right)}^2} - 2ab - 2bc - 2ca} \right)^2} - 2{a^2}{b^2} - 2{b^2}{c^2} - 2{c^2}{a^2} \cr
& = 2{\left( {ab + bc + ca} \right)^2} + 2\left( {{a^2}bc + {b^2}ac + {c^2}ab} \right)\;\;\;\;\left( {do\;a + b + c = 0} \right) \cr
& = 2{\left( {ab + bc + ca} \right)^2}\;\;\;\;\;\;\;\left( {do\;a + b + c = 0} \right) \cr
& co:0 = \left( {{a^2} + {b^2} + {c^2}} \right)\left( {a + b + c} \right) = {a^3} + {b^3} + {c^3} + {a^2}\left( {b + c} \right) + {b^2}\left( {c + a} \right) + {c^2}\left( {a + b} \right) \cr
& \leftrightarrow - \left( {{a^3} + {b^3} + {c^3}} \right) = {a^2}\left( {b + c} \right) + {b^2}\left( {c + a} \right) + {c^2}\left( {a + b} \right)\;\left( 2 \right) \cr
& co:{a^2} + {b^2} + {c^2} = {\left( {a + b + c} \right)^2} - 2\left( {ab + bc + ca} \right) = - 2\left( {ab + bc + ca} \right)\;\left( 3 \right) \cr
& thay\;\left( 1 \right)\;\left( 2 \right)\;\left( 3 \right)\;vao\;M \cr
& \to M = 2{\left( {ab + bc + ca} \right)^2} + 4\left( {{a^2}\left( {b + c} \right) + {b^2}\left( {c + a} \right) + {c^2}\left( {a + b} \right)} \right) - 12\left( {ab + bc + ca} \right) + 12abc + 3 \cr
& ma:\;4\left( {{a^2}\left( {b + c} \right) + {b^2}\left( {c + a} \right) + {c^2}\left( {a + b} \right)} \right) + 12abc = 4ab\left( {a + b + c} \right) + 4bc\left( {a + b + c} \right) + 4ca\left( {a + b + c} \right) = 0 \cr
& \to M = 2{\left( {ab + bc + ca} \right)^2} - 12\left( {ab + bc + ca} \right) + 3 \cr
& dat\;t = ab + bc + ca \cr
& co:\;0 = {\left( {a + b + c} \right)^2} \to t = {{ - \left( {{a^2} + {b^2} + {c^2}} \right)} \over 2} \le 0 \cr
& do\;a + b + c = 0 \to xet\;3\;truong\;hop: \cr
& TH1:a = b = c = 0 \to t = 0 \cr
& TH2:\;trong\;3\;so\;a,b,c\;co\;1\;so\;am\;va\;2\;so\;khong\;am\;do\;tinh\;doi\;xung\; \to {\rm{gia}}\;su\;so\;am\;la\;a\;so\;khong\;am\;la\;b,c \cr
& do\; - 1 \le a,b,c \le 1 \to {a^2} + {b^2} + {c^2} \le \left| a \right| + \left| b \right| + \left| c \right| = - a + b + c = \left( {a + b + c} \right) - 2a = - 2a \le 2\;\left( {do - 1 \le a} \right) \cr
& \to t = {{ - \left( {{a^2} + {b^2} + {c^2}} \right)} \over 2} \ge - 1 \cr
& dau = \leftrightarrow ... \cr
& TH3:trong\;3\;so\;co\;2\;so\;am\;va\;1\;so\;khong\;am\;tuong\;tu \to t \ge - 1 \cr
& tu\;cac\;dieu\;kien\;tren \to - 1 \le t \le 0 \cr
& co\;M = 2{t^2} - 12t + 3 \cr
& \to tim\;max\;min\;de\;roi \cr} $
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn