Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {x - y} \right)^2} \ge 0 \Leftrightarrow {x^2} + {y^2} - 2xy \ge 0\\
\Leftrightarrow {x^2} + 2xy + {y^2} \ge 4xy\\
\Leftrightarrow {\left( {x + y} \right)^2} \ge 4xy\\
\Leftrightarrow \frac{{x + y}}{{xy}} \ge \frac{4}{{x + y}}\\
\Leftrightarrow \frac{1}{x} + \frac{1}{y} \ge \frac{4}{{x + y}},\,\,\,\,\forall x,y > 0
\end{array}\)
Dấu '=' xảy ra khi và chỉ khi x=y
Áp dụng BĐT trên ta có:
\(\begin{array}{l}
\frac{1}{{2xy + xz + yz}} + \frac{1}{{xy + 2yz + zx}} + \frac{1}{{xy + yz + 2zx}}\\
= \frac{1}{4}\left( {\frac{4}{{\left( {xy + yz} \right) + \left( {xy + zx} \right)}} + \frac{4}{{\left( {xy + yz} \right) + \left( {yz + zx} \right)}} + \frac{4}{{\left( {xy + zx} \right) + \left( {yz + zx} \right)}}} \right)\\
\le \frac{1}{4}.\left( {\frac{1}{{xy + yz}} + \frac{1}{{xy + zx}} + \frac{1}{{xy + yz}} + \frac{1}{{yz + zx}} + \frac{1}{{xy + zx}} + \frac{1}{{yz + zx}}} \right)\\
\le \frac{1}{{16}}.\left( {\frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{xy}} + \frac{1}{{zx}} + \frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{yz}} + \frac{1}{{zx}} + \frac{1}{{xy}} + \frac{1}{{zx}} + \frac{1}{{yz}} + \frac{1}{{zx}}} \right)\\
= \frac{1}{{16}}.4.\left( {\frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{zx}}} \right)\\
= \frac{1}{4}.\frac{{x + y + z}}{{xyz}} = \frac{1}{{xyz}}
\end{array}\)
Dấu '=' xảy ra khi và chỉ khi \(x = y = z = \frac{4}{3}\)
