bài hơi dài tý :d
$\eqalign{
& P = {{{a^2}} \over {a + 2{b^2}}} + {{{b^2}} \over {b + 2{c^2}}} + {{{c^2}} \over {c + 2{a^2}}} \cr
& con\;nay\;dung\;ki\;thua\;cosi\;nguoc\;dau: \cr
& {{{a^2}} \over {a + 2{b^2}}} = a - {{2a{b^2}} \over {a + 2{b^2}}} \ge a - {{2a{b^2}} \over {3\root 3 \of {a{b^4}} }} = a - {{2\root 3 \of {{a^2}{b^2}} } \over 3}\;(do:\;a + 2{b^2} = a + {b^2} + {b^2} \ge 3\root 3 \of {a{b^4}} \to ...) \cr
& tuong\;tu: \cr
& {{{b^2}} \over {b + 2{c^2}}} \ge b - {{2\root 3 \of {{b^2}{c^2}} } \over 3} \cr
& {{{c^2}} \over {c + 2{a^2}}} \ge c - {{2\root 3 \of {{c^2}{a^2}} } \over 3} \cr
& \to P \ge a + b + c - {2 \over 3}\left( {\root 3 \of {{a^2}{b^2}} + \root 3 \of {{b^2}{c^2}} + \root 3 \of {{c^2}{a^2}} } \right) \cr
& co: \cr
& 3\left( {a + b + c} \right) = \left( {a + b + c} \right)\left( {\sqrt a + \sqrt b + \sqrt c } \right) = a\sqrt a + a\sqrt b + a\sqrt c + b\sqrt a + b\sqrt b + b\sqrt c + c\sqrt a + c\sqrt b + c\sqrt c \cr
& \cos i: \cr
& {{a\sqrt a } \over 2} + {{b\sqrt b } \over 2} + a\sqrt b + b\sqrt a + {3 \over 8} \cr
& = \underbrace {{{a\sqrt a } \over 8} + ... + {{a\sqrt a } \over 8}}_{4\;so} + \underbrace {{{b\sqrt b } \over 8} + ... + {{b\sqrt b } \over 8}}_{4\;so} + \underbrace {{{a\sqrt b } \over 8} + ... + {{a\sqrt b } \over 8}}_{8\;so} + \underbrace {{{b\sqrt a } \over 8} + ... + {{b\sqrt a } \over 8}}_{8\;so} + {1 \over 8} + {1 \over 8} + {1 \over 8} \ge 27\root {27} \of {{{{a^{18}}{b^{18}}} \over {{8^{27}}}}} = {{27\root 3 \of {{a^2}{b^2}} } \over 8} \cr
& tuong\;tu: \cr
& {{b\sqrt b } \over 2} + {{c\sqrt c } \over 2} + b\sqrt c + c\sqrt b + {3 \over 8} \ge {{27\root 3 \of {{b^2}{c^2}} } \over 8} \cr
& {{c\sqrt c } \over 2} + {{a\sqrt a } \over 2} + c\sqrt a + a\sqrt c + {3 \over 8} \ge {{27\root 3 \of {{c^2}{a^2}} } \over 8} \cr
& \to 3\left( {a + b + c} \right) + {9 \over 8} = a\sqrt a + a\sqrt b + a\sqrt c + b\sqrt a + b\sqrt b + b\sqrt c + c\sqrt a + c\sqrt b + c\sqrt c + {9 \over 8} \ge {{27} \over 8}\left( {\root 3 \of {{a^2}{b^2}} + \root 3 \of {{b^2}{c^2}} + \root 3 \of {{c^2}{a^2}} } \right) \cr
& \leftrightarrow {{16} \over {27}}\left( {a + b + c} \right) + {2 \over 9} \ge {2 \over 3}\left( {\root 3 \of {{a^2}{b^2}} + \root 3 \of {{b^2}{c^2}} + \root 3 \of {{c^2}{a^2}} } \right) \cr
& \to P \ge a + b + c - {2 \over 3}\left( {\root 3 \of {{a^2}{b^2}} + \root 3 \of {{b^2}{c^2}} + \root 3 \of {{c^2}{a^2}} } \right) = {{11} \over {27}}\left( {a + b + c} \right) - {2 \over 9} + {{16} \over {27}}\left( {a + b + c} \right) + {2 \over 9} - {2 \over 3}\left( {\root 3 \of {{a^2}{b^2}} + \root 3 \of {{b^2}{c^2}} + \root 3 \of {{c^2}{a^2}} } \right) \cr
& \ge {{11} \over {27}}\left( {a + b + c} \right) - {2 \over 9} \ge 1 \cr
& dau = \leftrightarrow a = b = c = 1 \cr} $
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