Ta có: a,b,c > 1 và x, y, z > 0 nên \({a^x};{b^y};{c^z};\sqrt[3]{{abc}} > 1\)

Do đó:

\({a^x} = {b^y} = {c^z} = \sqrt[3]{{abc}} \Leftrightarrow \left\{ \begin{array}{l} x = \frac{1}{3}\left( {1 + {{\log }_a}b + {{\log }_a}c} \right)\\ y = \frac{1}{3}\left( {{{\log }_b}a + 1 + {{\log }_b}c} \right)\\ z = \frac{1}{3}\left( {{{\log }_c}a + {{\log }_c}b + 1} \right) \end{array} \right.\).

Khi đó, ta có:

\(\begin{array}{l} P = x + y + z = \frac{1}{3}\left( {1 + {{\log }_a}b + {{\log }_a}c + {{\log }_b}a + 1 + {{\log }_b}c + {{\log }_c}a + {{\log }_c}b + 1} \right)\\ = \frac{1}{3}.\left( {3 + {{\log }_a}b + {{\log }_a}c + {{\log }_b}a + {{\log }_b}c + {{\log }_c}a + {{\log }_c}b} \right)\\ = \frac{1}{3}.\left( {3 + {{\log }_a}b + {{\log }_b}c + {{\log }_c}a + {{\log }_a}c + {{\log }_c}b + {{\log }_b}a} \right) \end{array}\)

Mặt khác a,b,c > 1 nên \({\log _a}b,{\log _b}c,{\log _c}a,{\log _a}c,{\log _c}b,{\log _b}a > 0\)

Suy ra: \(P \ge \frac{1}{3}\left( {3 + 3\sqrt[3]{{{{\log }_a}b.{{\log }_b}c.{{\log }_c}a}} + 3\sqrt[3]{{{{\log }_a}c.{{\log }_c}b.{{\log }_b}a}}} \right) = 3\).

Dấu "=" xảy ra khi: \(\left\{ \begin{array}{l} {\log _a}b = {\log _b}c = {\log _c}a\\ {\log _a}c = {\log _c}b = {\log _b}a\\ {a^x} = {b^y} = {c^z} = \sqrt[3]{{abc}} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {\log _a}b = {\log _b}c = {\log _c}a\\ \frac{1}{{{{\log }_c}a}} = \frac{1}{{{{\log }_b}c}} = \frac{1}{{{{\log }_a}b}}\\ {a^x} = {b^y} = {c^z} = \sqrt[3]{{abc}} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = b = c\\ x = y = z = 1 \end{array} \right.\).

Vậy \({\mathop{\rm minP}\nolimits} = 3 \in \left( {2;4} \right).\)