Cho hai số thực a, b thỏa mãn và biểu thức có giá trị nhỏ nhất. Tính .

* Hướng dẫn giải

Ta có: \(4{b^3} - 3b + 1 = \left( {b + 1} \right){\left( {2b - 1} \right)^2} \ge 0;\forall b \in \left( {\frac{1}{3};1} \right).\)

Suy ra: \(3b - 1 \le 4{b^3} \Rightarrow {\log _a}\left( {\frac{{3b - 1}}{{4{a^3}}}} \right) \ge {\log _a}\left( {\frac{{4{b^3}}}{{4{a^3}}}} \right),\) do \(a \in \left( {\frac{1}{3};1} \right)\).

\( \Rightarrow P \ge 3{\log _a}\left( {\frac{b}{a}} \right) + 12\log _{\frac{b}{a}}^2a = 3\left[ {\frac{1}{2}{{\log }_a}\left( {\frac{b}{a}} \right) + \frac{1}{2}{{\log }_a}\left( {\frac{b}{a}} \right) + \frac{4}{{\log _a^2\left( {\frac{b}{a}} \right)}}} \right]\) .

\( \ge 3.3\sqrt[3]{{\frac{1}{2}{{\log }_a}\left( {\frac{b}{a}} \right)\frac{1}{2}{{\log }_a}\left( {\frac{b}{a}} \right)\frac{4}{{\log _a^2\left( {\frac{b}{a}} \right)}}}} = 9\)

\({P_{\min }} = 9 \Leftrightarrow \left\{ \begin{array}{l} b = \frac{1}{2}\\ \frac{1}{2}{\log _a}\left( {\frac{b}{a}} \right) = \frac{4}{{\log _a^2\left( {\frac{b}{a}} \right)}} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} b = \frac{1}{2}\\ {\log _a}\left( {\frac{b}{a}} \right) = 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} b = \frac{1}{2}\\ \frac{b}{a} = {a^2} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} b = \frac{1}{2}\\ a = \frac{1}{{\sqrt[3]{2}}} \end{array} \right.\)

Vậy \(\frac{b}{a} = \frac{1}{{\sqrt[3]{4}}}\)