Đáp án:
Giải thích các bước giải:
Sử dụng bđt \(\frac{1}{x} + \frac{1}{y} \ge \frac{4}{{x + y}} \Leftrightarrow \frac{1}{{x + y}} \le \frac{1}{4}\left( {\frac{1}{x} + \frac{1}{y}} \right)\)
Ta có:
\(\begin{array}{l}\frac{a}{{ca + 1}} = \frac{a}{{ca + abc}} = \frac{1}{{c + bc}} \le \frac{1}{4}\left( {\frac{1}{c} + \frac{1}{{bc}}} \right)\\\frac{b}{{ab + 1}} = \frac{b}{{ab + abc}} = \frac{1}{{a + ac}} \le \frac{1}{4}\left( {\frac{1}{a} + \frac{1}{{ac}}} \right)\\\frac{c}{{bc + 1}} = \frac{c}{{bc + abc}} \le \frac{1}{{b + ab}} \le \frac{1}{4}\left( {\frac{1}{b} + \frac{1}{{ab}}} \right)\\ \Rightarrow P \le \frac{1}{4}\left( {\frac{1}{c} + \frac{1}{{bc}} + \frac{1}{a} + \frac{1}{{ac}} + \frac{1}{b} + \frac{1}{{ab}}} \right) = \frac{1}{4}\left( {ab + a + bc + b + ca + b} \right)\\ = \frac{1}{4}\left( {ab + bc + ca + a + b + c} \right)\end{array}\)
Mà \(ab + bc + ca \le {a^2} + {b^2} + {c^2}\) và:
\(a \le \frac{{{a^2} + 1}}{2};b \le \frac{{{b^2} + 1}}{2};c \le \frac{{{c^2} + 1}}{2} \Rightarrow a + b + c \le \frac{{{a^2} + {b^2} + {c^2} + 3}}{2}\)
Mà \({a^2} + {b^2} + {c^2} \ge 3\sqrt[3]{{{{\left( {abc} \right)}^2}}} = 3\) hay \(3 \le {a^2} + {b^2} + {c^2}\)
\( \Rightarrow a + b + c \le \frac{{{a^2} + {b^2} + {c^2} + {a^2} + {b^2} + {c^2}}}{2} = {a^2} + {b^2} + {c^2}\) \(\begin{array}{l} \Rightarrow \frac{1}{4}\left( {ab + bc + ca + a + b + c} \right) \le \frac{1}{4}\left( {{a^2} + {b^2} + {c^2} + {a^2} + {b^2} + {c^2}} \right)\\ \Rightarrow P \le \frac{1}{2}\left( {{a^2} + {b^2} + {c^2}} \right)\end{array}\)
