Giải thích các bước giải:
ĐK: $x,y,z>0$
Ta có:
$\begin{array}{l}
x + y + z + xy + yz + xz = 6xyz\\
\Leftrightarrow \dfrac{{x + y + z + xy + yz + xz}}{{xyz}} = 6\\
\Leftrightarrow \dfrac{1}{{yz}} + \dfrac{1}{{xz}} + \dfrac{1}{{xy}} + \dfrac{1}{z} + \dfrac{1}{x} + \dfrac{1}{y} = 6
\end{array}$
Mà lại có:
$\begin{array}{l}
A = \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} + \dfrac{1}{{{z^2}}}\\
\Rightarrow 3A + 3 = \left( {\dfrac{1}{{{x^2}}} + 1} \right) + \left( {\dfrac{1}{{{y^2}}} + 1} \right) + \left( {\dfrac{1}{{{z^2}}} + 1} \right) + \left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}}} \right) + \left( {\dfrac{1}{{{y^2}}} + \dfrac{1}{{{z^2}}}} \right) + \left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{{{z^2}}}} \right)\\
\Rightarrow 3A + 3 \ge 2\sqrt {\dfrac{1}{{{x^2}}}} + 2\sqrt {\dfrac{1}{{{y^2}}}} + 2\sqrt {\dfrac{1}{{{z^2}}}} + 2\sqrt {\dfrac{1}{{{x^2}}}.\dfrac{1}{{{y^2}}}} + 2\sqrt {\dfrac{1}{{{y^2}}}.\dfrac{1}{{{z^2}}}} + 2\sqrt {\dfrac{1}{{{x^2}}}.\dfrac{1}{{{z^2}}}} \\
\Rightarrow 3A + 3 \ge 2.\dfrac{1}{x} + 2.\dfrac{1}{y} + 2.\dfrac{1}{z} + 2.\dfrac{1}{{xy}} + 2.\dfrac{1}{{yz}} + 2.\dfrac{1}{{xz}}\\
\Rightarrow 3A + 3 \ge 2\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} + \dfrac{1}{{xy}} + \dfrac{1}{{yz}} + \dfrac{1}{{xz}}} \right)\\
\Rightarrow 3A + 3 \ge 2.6\\
\Rightarrow A \ge 3
\end{array}$
Dấu bằng xảy ra
$ \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{x} = 1\\
\dfrac{1}{y} = 1\\
\dfrac{1}{z} = 1\\
\dfrac{1}{x} = \dfrac{1}{y}\\
\dfrac{1}{y} = \dfrac{1}{z}\\
\dfrac{1}{x} = \dfrac{1}{z}
\end{array} \right. \Leftrightarrow x = y = z = 1$
Vậy ta có điều phải chứng minh.
