Xét các số thực a, b thỏa mãn điều kiện \( - \ln 3 var DOMA...

* Hướng dẫn giải

Ta có \({\left( {2b - 1} \right)^2}\left( {b + 1} \right) \ge 0 \Rightarrow 3b - 1 \le 4{b^3} \Rightarrow \frac{{3b - 1}}{4} \le {b^3}\). Vì \(\frac{1}{3} < b < a < 1\) nên \({\log _a}b > 1\).

Suy ra \(P \ge {\log _a}{b^3} + \frac{{12}}{{{{\left( {{{\log }_a}\frac{b}{a}} \right)}^2}}} - 3 \ge 3\left( {{{\log }_a}b - 1} \right) + \frac{{12}}{{{{\left( {{{\log }_a}b - 1} \right)}^2}}}\)

\( \ge \frac{3}{2}\left( {{{\log }_a}b - 1} \right) + \frac{3}{2}\left( {{{\log }_a}b - 1} \right) + \frac{{12}}{{{{\left( {{{\log }_a}b - 1} \right)}^2}}}\)

\( \ge 3.\sqrt[3]{{\frac{3}{2}\left( {{{\log }_a}b - 1} \right).\frac{3}{2}\left( {{{\log }_a}b - 1} \right).\frac{{12}}{{{{\left( {{{\log }_a}b - 1} \right)}^2}}}}} \ge 9\)

Dấu "=" xảy ra khi \(\left\{ \begin{array}{l} 2b - 1 = 0\\ \frac{3}{2}\left( {{{\log }_a}b - 1} \right) = \frac{{12}}{{{{\left( {{{\log }_a}b - 1} \right)}^2}}} \end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l} b = \frac{1}{2}\\ a = \frac{1}{{\sqrt[3]{2}}} \end{array} \right.\)