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- nguyenhongsonk612 yêu thích
Gửi bởi LsTinyBaby trong 08-12-2014 - 00:10
Gửi bởi LsTinyBaby trong 24-07-2013 - 17:07
Áp dụng Cauchy-Schwarz ta có được \[\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}} + \frac{1}{{2\sqrt[3]{{abc}}}} = \sum\limits_{cyc} {\frac{{{c^2}}}{{{c^2}\left( {a + b} \right)}}} + \frac{{{{\left( {\sqrt[3]{{abc}}} \right)}^2}}}{{2abc}} \ge \frac{{{{\left( {c + a + b + \sqrt[3]{{abc}}} \right)}^2}}}{{\sum\limits_{cyc} {{c^2}\left( {a + b} \right)} + 2abc}} = \frac{{{{\left( {a + b + c + \sqrt[3]{{abc}}} \right)}^2}}}{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}}\]
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