Xét ba số \(x;y;z \ge 2\) thỏa mãn \(4xyz = 9\left( {x + y + z} \right) + 27\)

Giải. Ta có \(\sqrt 5 Q\) = \(\frac{{\sqrt {5\left( {x - 2} \right)(x\_ + 2} }}{x} + \frac{{\sqrt {5\left( {y - 2} \right)\left( {y + 2} \right)} }}{y} + \frac{{\sqrt {5\left( {z - 2} \right)\left( {z + 2} \right)} }}{z}\)

\(\sqrt 5 Q \le \)\(\frac{{5\left( {x - 2} \right) + x + 2}}{{2x}} + \frac{{5\left( {y - 2} \right) + y + 2}}{{2y}} + \frac{{5\left( {z - 2} \right) + z + 2}}{{2z}}\)

 \( \Leftrightarrow \sqrt 5 Q \le \)\(\frac{{6x - 8}}{{2x}} + \frac{{6y - 8}}{{2y}} + \frac{{6z - 8}}{{2z}}\) = 9 – 4\(1\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)\)

Từ \(4xyz = 9\left( {x + y + z} \right) + 27 \Leftrightarrow \) 4 = 9\(\left( {\frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{xz}}} \right) + \frac{{27}}{{xyz}}\)\( \le \) 3\({\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)^2} + {\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)^3}\)

Đặt \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = t\)

Ta có

      \({t^3} + 3{t^2} - 4 \ge 0 \Leftrightarrow {t^3} - {t^2} + 4{t^2} - 4t + 4t - 4 \ge 0\)

                                     \( \Leftrightarrow \left( {t - 1} \right){\left( {t - 2} \right)^2} \ge 0\)

                                                                                     \( \Leftrightarrow t \ge 1\)

Suy ra \(\sqrt 5 Q \le 9 - 4.1 = 5 \Leftrightarrow Q \le \sqrt 5 \)

Vậy \(MaxQ = \sqrt 5 \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x,y,z \ge 2;4xyz = 9\left( {x + y + z} \right) + 27\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\{5\left( {x - 2} \right) = x + 2;5\left( {y - 2} \right) = y + 2;5\left( {z - 2} \right) = z + 2}\\{\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{array}} \right.\)

                                                                              \( \Leftrightarrow x = y = z = 3\)