Đáp án:
\[{P_{\max }} = 1\]
Giải thích các bước giải:
Áp dụng BĐT sau:
\[\begin{array}{l}
{\left( {a - b} \right)^2} \ge 0\\
\Leftrightarrow {a^2} + {b^2} - 2ab \ge 0\\
\Leftrightarrow {a^2} + 2ab + {b^2} \ge 4ab\\
\Leftrightarrow {\left( {a + b} \right)^2} \ge 4ab\\
\Leftrightarrow \frac{{a + b}}{{ab}} \ge \frac{4}{{a + b}}\\
\Leftrightarrow \frac{1}{a} + \frac{1}{b} \ge \frac{4}{{a + b}}
\end{array}\]
Ta có:
\(\begin{array}{l}
P = \frac{1}{{2x + y + z}} + \frac{1}{{x + 2y + z}} + \frac{1}{{x + y + 2z}}\\
= \frac{1}{4}\left( {\frac{4}{{\left( {x + y} \right) + \left( {x + z} \right)}} + \frac{4}{{\left( {x + y} \right) + \left( {y + z} \right)}} + \frac{4}{{\left( {x + z} \right) + \left( {y + z} \right)}}} \right)\\
\le \frac{1}{4}.\left( {\frac{1}{{x + y}} + \frac{1}{{x + z}} + \frac{1}{{x + y}} + \frac{1}{{y + z}} + \frac{1}{{z + x}} + \frac{1}{{z + y}}} \right)\\
= \frac{1}{{16}}\left( {\frac{4}{{x + y}} + \frac{4}{{x + z}} + \frac{4}{{x + y}} + \frac{4}{{y + z}} + \frac{4}{{z + x}} + \frac{4}{{z + y}}} \right)\\
\le \frac{1}{{16}}\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{x} + \frac{1}{z} + \frac{1}{x} + \frac{1}{y} + \frac{1}{y} + \frac{1}{z} + \frac{1}{z} + \frac{1}{x} + \frac{1}{z} + \frac{1}{y}} \right)\\
= \frac{1}{{16}}.4.\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) = 1
\end{array}\)
Dấu '=' xảy ra khi và chỉ khi \(x = y = z = \frac{3}{4}\)
Vậy \({P_{\max }} = 1\)
