$\eqalign{
& \cos i: \cr
& {\left( {{{a + 2b} \over {a + 2c}}} \right)^3} + 1 + 1 \ge {{3\left( {a + 2b} \right)} \over {a + 2c}} \cr
& {\left( {{{b + 2c} \over {b + 2a}}} \right)^3} + 1 + 1 \ge {{3\left( {b + 2c} \right)} \over {b + 2a}} \cr
& {\left( {{{c + 2a} \over {c + 2b}}} \right)^3} + 1 + 1 \ge {{3\left( {c + 2a} \right)} \over {c + 2b}} \cr
& \to S + 6 = {\left( {{{a + 2b} \over {a + 2c}}} \right)^3} + {\left( {{{b + 2c} \over {b + 2a}}} \right)^3} + {\left( {{{c + 2a} \over {c + 2b}}} \right)^3} + 6 \ge 3\left( {{{a + 2b} \over {a + 2c}} + {{b + 2c} \over {b + 2a}} + {{c + 2a} \over {c + 2b}}} \right) \cr
& ma: \cr
& P = {{a + 2b} \over {a + 2c}} + {{b + 2c} \over {b + 2a}} + {{c + 2a} \over {c + 2b}} + 1 + 1 + 1 = \left( {2a + 2b + 2c} \right)\left( {{1 \over {a + 2c}} + {1 \over {b + 2a}} + {1 \over {c + 2b}}} \right) \cr
& \cos i:{1 \over {a + 2c}} + {1 \over {b + 2a}} + {1 \over {c + 2b}} \ge 3\sqrt {{1 \over {\left( {a + 2c} \right)\left( {b + 2a} \right)\left( {c + 2b} \right)}}} \ge {3 \over {\left( {{{a + 2c + b + 2a + c + 2b} \over 3}} \right)}} = {3 \over {a + b + c}} \cr
& \to P \ge \left( {2a + 2b + 2c} \right)*{3 \over {a + b + c}} = 6 \cr
& \to P - 3 = {{a + 2b} \over {a + 2c}} + {{b + 2c} \over {b + 2a}} + {{c + 2a} \over {c + 2b}} \ge 3 \cr
& \to S + 6 \ge 3*3 = 9 \cr
& \leftrightarrow S \ge 3 \cr
& dau = \leftrightarrow a = b = c \cr} $
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