Lời giải
Trước hết ta đi chứng minh bổ đề sau: \[\frac{1}{{2 + a}} + \frac{1}{{2 + b}} + \frac{1}{{2 + c}} \le 1\]
Ta có:
$$\begin{array}{l}
\frac{1}{{2 + a}} + \frac{1}{{2 + b}} + \frac{1}{{2 + c}} \le 1\\
\Leftrightarrow 1 - \frac{2}{{2 + a}} + 1 - \frac{2}{{2 + b}} + 1 - \frac{2}{{2 + c}} \ge 1\\
\Leftrightarrow \frac{a}{{2 + a}} + \frac{b}{{2 + b}} + \frac{c}{{2 + c}} \ge 1
\end{array}$$
\frac{1}{{2 + a}} + \frac{1}{{2 + b}} + \frac{1}{{2 + c}} \le 1\\
\Leftrightarrow 1 - \frac{2}{{2 + a}} + 1 - \frac{2}{{2 + b}} + 1 - \frac{2}{{2 + c}} \ge 1\\
\Leftrightarrow \frac{a}{{2 + a}} + \frac{b}{{2 + b}} + \frac{c}{{2 + c}} \ge 1
\end{array}$$
Tồn tại các số thực $x,y,z$ sao cho $a = \frac{x}{y};\,\,b = \frac{y}{z};\,\,c = \frac{z}{x}$
Ta cần chứng minh.
$$\begin{array}{l}
\frac{{\frac{x}{y}}}{{2 + \frac{x}{y}}} + \frac{{\frac{y}{z}}}{{2 + \frac{y}{z}}} + \frac{{\frac{z}{x}}}{{2 + \frac{z}{x}}} = \frac{x}{{2y + x}} + \frac{y}{{2z + y}} + \frac{z}{{2x + z}} \ge 1\\
\Leftrightarrow \frac{{{x^2}}}{{2xy + {x^2}}} + \frac{{{y^2}}}{{2yz + {y^2}}} + \frac{{{z^2}}}{{2zx + {z^2}}} \ge 1
\end{array}$$
\frac{{\frac{x}{y}}}{{2 + \frac{x}{y}}} + \frac{{\frac{y}{z}}}{{2 + \frac{y}{z}}} + \frac{{\frac{z}{x}}}{{2 + \frac{z}{x}}} = \frac{x}{{2y + x}} + \frac{y}{{2z + y}} + \frac{z}{{2x + z}} \ge 1\\
\Leftrightarrow \frac{{{x^2}}}{{2xy + {x^2}}} + \frac{{{y^2}}}{{2yz + {y^2}}} + \frac{{{z^2}}}{{2zx + {z^2}}} \ge 1
\end{array}$$
Theo Svacxo ta có:
$$\begin{array}{l}
\frac{{{x^2}}}{{2xy + {x^2}}} + \frac{{{y^2}}}{{2yz + {y^2}}} + \frac{{{z^2}}}{{2zx + {z^2}}} \ge \frac{{{{\left( {x + y + z} \right)}^2}}}{{2xy + {x^2} + 2yz + {y^2} + 2zx + {z^2}}}\\
= \frac{{{{\left( {x + y + z} \right)}^2}}}{{{{\left( {x + y + z} \right)}^2}}} = 1
\end{array}$$
\frac{{{x^2}}}{{2xy + {x^2}}} + \frac{{{y^2}}}{{2yz + {y^2}}} + \frac{{{z^2}}}{{2zx + {z^2}}} \ge \frac{{{{\left( {x + y + z} \right)}^2}}}{{2xy + {x^2} + 2yz + {y^2} + 2zx + {z^2}}}\\
= \frac{{{{\left( {x + y + z} \right)}^2}}}{{{{\left( {x + y + z} \right)}^2}}} = 1
\end{array}$$
Dấu bằng xảy ra khi: $x = y = z \Leftrightarrow a = b = c$
Áp dụng bất đẳng thức trên ta có:
\[\begin{array}{l}\frac{1}{{2 + a}} + \frac{1}{{2 + b}} \le 1 - \frac{1}{{2 + c}}\\
\Leftrightarrow \frac{1}{{2 + a}} + \frac{1}{{2 + b}} + \frac{2}{{\sqrt {3\left( {2 + c} \right)} }} \le 1 - \frac{1}{{2 + c}} + \frac{2}{{\sqrt {3\left( {2 + c} \right)} }}\\
\Leftrightarrow T \le 1 - \frac{1}{{2 + c}} + \frac{2}{{\sqrt {3\left( {2 + c} \right)} }}\\
\Leftrightarrow T \le - \left( {\frac{1}{{2 + c}} - \frac{2}{{\sqrt {3\left( {2 + c} \right)} }} + \frac{1}{3}} \right) + \frac{4}{3}\\
\Leftrightarrow T \le - {\left( {\frac{1}{{\sqrt {2 + c} }} - \frac{1}{{\sqrt 3 }}} \right)^2} + \frac{4}{3} \le \frac{4}{3}
\end{array}\]
Dấu bằng xảy ra khi: $\frac{1}{{\sqrt {2 + c} }} = \frac{1}{{\sqrt 3 }} \Leftrightarrow c = 1$
Vậy giá trị lớn nhất của $P = \frac{4}{3}$ khi $a=b=c=1$
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