$\eqalign{
& \cos i: \cr
& \left( {{{{a^4}} \over {b + 3c}} + {{b + 3c} \over {16}}} \right) + \left( {{{{b^4}} \over {c + 3a}} + {{c + 3a} \over {16}}} \right) + \left( {{{{c^4}} \over {a + 3b}} + {{a + 3b} \over {16}}} \right) \ge {{{a^2}} \over 2} + {{{b^2}} \over 2} + {{{c^2}} \over 2} \cr
& \to VT \ge {{{a^2} + {b^2} + {c^2}} \over 2} - {{a + b + c} \over 4} \cr
& ma\;cosi: \cr
& {{\left( {{a^2} + 1 + {b^2} + 1 + {c^2} + 1} \right)} \over 8} \ge {{2a + 2b + 2c} \over 8} \to {{{a^2} + {b^2} + {c^2}} \over 8} \ge {{a + b + c} \over 4} - {3 \over 8} \cr
& 2\left( {{a^2} + {b^2} + {c^2}} \right) = {a^2} + {b^2} + {b^2} + {c^2} + {c^2} + {a^2} \ge 2ab + 2bc + 2ca \ge 6 \cr
& \to {a^2} + {b^2} + {c^2} \ge 3 \cr
& \to {{{a^2} + {b^2} + {c^2}} \over 2} - {{a + b + c} \over 4} = {{{a^2} + {b^2} + {c^2}} \over 8} - {{a + b + c} \over 4} + {{3\left( {{a^2} + {b^2} + {c^2}} \right)} \over 8} \ge {{a + b + c} \over 4} - {3 \over 8} - {{a + b + c} \over 4} - {{3*3} \over 8} = {3 \over 4} \cr
& dau = \leftrightarrow a = b = c = 1 \cr} $