$\eqalign{
& do\;tam\;giac\;khong\;nhon \to {\text{co}}\;mot\;goc\;tu\;\left( {hoac\;vuong} \right) \cr
& goi\;a\;la\;canh\;doi\;dien\;voi\;goc\;tu \cr
& \to de\;dang\;cm\;{a^2} \geqslant {b^2} + {c^2} \cr
& \cos i: \cr
& {a^2} + {b^2} + {c^2} = \frac{{{a^2}}}{2} + \frac{{{a^2}}}{2} + {b^2} + {c^2} \geqslant 4\root 4 \of {\frac{{{a^4}{b^2}{c^2}}}{4}} \cr
& \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{1}{2}\left( {\frac{2}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} + \frac{1}{{{c^2}}}} \right) \geqslant \frac{1}{2}*5\root 5 \of {\frac{2}{{{a^2}{b^4}{c^4}}}} \cr
& \to \left( {{a^2} + {b^2} + {c^2}} \right)\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} \right) \geqslant 10*\root 4 \of {\frac{{{a^4}{b^2}{c^2}}}{4}} *\root 5 \of {\frac{2}{{{a^2}{b^4}{c^4}}}} = 10\root {20} \of {\frac{{{a^{20}}{b^{10}}{c^{10}}}}{{{4^5}}}*\frac{{{2^4}}}{{{a^8}{b^{16}}{c^{16}}}}} = 10\root {20} \of {\frac{{{a^{12}}}}{{{2^6}{b^6}{c^6}}}} \cr
& ma\;{a^2} \geqslant {b^2} + {c^2} \geqslant 2bc \to {a^{12}} \geqslant {2^6}{b^6}{c^6} \cr
& \to 10\root {20} \of {\frac{{{a^{12}}}}{{{2^6}{b^6}{c^6}}}} \geqslant 10 \cr
& dau = \leftrightarrow {a^2} = 2{b^2} = 2{c^2}\;hay\;tam\;giac\;vuong\;can \cr} $
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