Đáp án đúng:

Phương pháp giải:

Sử dụng bất đẳng thức Cauchy – Schwarz Giải chi tiết:Đặt \(P = \frac{{{a^2}}}{{5{a^2} + {{\left( {b + c} \right)}^2}}} + \frac{{{b^2}}}{{5{b^2} + {{\left( {c + a} \right)}^2}}} + \frac{{{c^2}}}{{5{c^2} + {{\left( {a + b} \right)}^2}}}\)Áp dụng bất đẳng thức Cauchy – Schwarz  ta có:\(\begin{array}{l}\left[ {\left( {{a^2} + {b^2} + {c^2}} \right) + \left( {2{a^2} + bc} \right) + \left( {2{a^2} + bc} \right)} \right].\left( {\frac{1}{{{a^2} + {b^2} + {c^2}}} + \frac{1}{{2{a^2} + bc}} + \frac{1}{{2{a^2} + bc}}} \right) \ge 9\\\frac{{9{a^2}}}{{5{b^2} + {{\left( {b + c} \right)}^2}}} = \frac{{9{a^2}}}{{\left( {{a^2} + {b^2} + {c^2}} \right) + \left( {2{a^2} + bc} \right) + \left( {2{a^2} + bc} \right)}} \le {a^2}\left( {\frac{1}{{{a^2} + {b^2} + {c^2}}} + \frac{2}{{2{a^2} + bc}}} \right)\end{array}\)Chứng minh tương tự, ta được:\(\begin{array}{l}\frac{{9{b^2}}}{{5{b^2} + {{\left( {c + a} \right)}^2}}} \le {b^2}\left( {\frac{1}{{{a^2} + {b^2} + {c^2}}} + \frac{2}{{2{b^2} + ac}}} \right)\\\frac{{9{c^2}}}{{5{c^2} + {{\left( {a + b} \right)}^2}}} \le {c^2}\left( {\frac{1}{{{a^2} + {b^2} + {c^2}}} + \frac{2}{{2{c^2} + ab}}} \right)\end{array}\)Khi đó ta có:\(\frac{{9{a^2}}}{{5{a^2} + {{\left( {b + c} \right)}^2}}} + \frac{{9{b^2}}}{{5{b^2} + {{\left( {c + a} \right)}^2}}} + \frac{{9{c^2}}}{{5{c^2} + {{\left( {a + b} \right)}^2}}} \le 1 + \left( {\frac{{2{a^2}}}{{2{a^2} + bc}} + \frac{{2{b^2}}}{{2{b^2} + ca}} + \frac{{2{c^2}}}{{2{c^2} + ab}}} \right)\)Suy ra\(9P \le 4 - \left( {\frac{{bc}}{{2{a^2} + bc}} + \frac{{ca}}{{2{b^2} + ca}} + \frac{{ab}}{{2{c^2} + ab}}} \right)\)Ta có :\(\frac{{bc}}{{2{a^2} + bc}} + \frac{{ca}}{{2{b^2} + ca}} + \frac{{ab}}{{2{c^2} + ab}} = \frac{{{b^2}{c^2}}}{{2{a^2}bc + {b^2}{c^2}}} + \frac{{{c^2}{a^2}}}{{2a{b^2}c + {c^2}{a^2}}} + \frac{{{a^2}{b^2}}}{{2ab{c^2} + {a^2}{b^2}}}\)Áp dụng bất đẳng thức (2) ta được:\(\frac{{{b^2}{c^2}}}{{2{a^2}bc + {b^2}{c^2}}} + \frac{{{c^2}{a^2}}}{{2a{b^2}c + {c^2}{a^2}}} + \frac{{{a^2}{b^2}}}{{2ab{c^2} + {a^2}{b^2}}} \ge \frac{{{{\left( {ab + bc + ca} \right)}^2}}}{{{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} + 2{a^2}bc + 2a{b^2}c + 2ab{c^2}}} = \frac{{{{\left( {ab + bc + ca} \right)}^2}}}{{{{\left( {ab + bc + ca} \right)}^2}}} = 1\)Vậy \(9P \le 4 - 1 \Rightarrow P \le \frac{1}{3}\). Dấu  xảy ra \( \Leftrightarrow a = b = c\).