$\eqalign{
& \cos i: \cr
& {a^2} + {1 \over {{a^2}}} = {a^2} + \underbrace {{1 \over {16{a^2}}} + ... + {1 \over {16{a^2}}}}_{16\;so} \ge 17\root {17} \of {{1 \over {{{16}^{16}}{a^{30}}}}} \cr
& tuong\;tu \cr
& {b^2} + {1 \over {{b^2}}} \ge 17\root {17} \of {{1 \over {{{16}^{16}}{b^{30}}}}} \cr
& \to \root 3 \of {{a^2} + {1 \over {{a^2}}}} + \root 3 \of {{b^2} + {1 \over {{b^2}}}} \ge \root 3 \of {17\root {17} \of {{1 \over {{{16}^{16}}{a^{30}}}}} } + \root 3 \of {17\root {17} \of {{1 \over {{{16}^{16}}{b^{30}}}}} } \cr
& = \root {51} \of {{{{{17}^{17}}} \over {{{16}^{16}}{a^{30}}}}} + \root {51} \of {{{{{17}^{17}}} \over {{{16}^{16}}{b^{30}}}}} \cr
& \cos i: \cr
& \root {51} \of {{{{{17}^{17}}} \over {{{16}^{16}}{a^{30}}}}} + \root {51} \of {{{{{17}^{17}}} \over {{{16}^{16}}{b^{30}}}}} \ge 2\sqrt {\root {51} \of {{{{{17}^{17}}} \over {{{16}^{16}}{a^{30}}}}} *\root {51} \of {{{{{17}^{17}}} \over {{{16}^{16}}{b^{30}}}}} } = 2\root {102} \of {{{{{17}^{34}}} \over {{{16}^{32}}{a^{30}}{b^{30}}}}} \left( {ma\;1 = a + b \ge {\rm{2}}\sqrt {ab} \to {1 \over {ab}} \ge 4} \right) \cr
& \ge 2\root {102} \of {{{{{17}^{34}}*{4^{30}}} \over {{{16}^{32}}}}} = 2\root {102} \of {{{{{17}^{34}}} \over {{4^{34}}}}} = 2\root 3 \of {{{17} \over 4}} \cr
& dau\; = \leftrightarrow a = b = {1 \over 2} \cr} $
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