Cho các số thực a,b thỏa mãn điều kiện 0

Đáp án D

Ta có \(9{b^2} - 12b + 4 = {\left( {3b - 2} \right)^2} \ge 0 \Rightarrow \frac{{4\left( {3b - 1} \right)}}{9} \le {b^2}\)

\( \Rightarrow P \ge {\log _a}{b^2} + 8\log _{\frac{b}{a}}^2a = 2{\log _a}b + 8{\left( {\frac{1}{{{{\log }_a}\frac{b}{a}}}} \right)^2} = 2{\log _a}b + \frac{8}{{{{\left( {{{\log }_a}b - 1} \right)}^2}}}.\)

Đặt \(t = {\log _a}b - 1 > 0 \Rightarrow P \ge 2\left( {t + 1} \right) + \frac{8}{{{t^2}}} = t + t + \frac{8}{{{t^2}}} \ge 3\sqrt[3]{{t.t.\frac{8}{{{t^2}}}}} + 2 = 8.\)

Dấu “=” xảy ra \( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{b = \frac{2}{3}}\\{t = \frac{8}{{{t^2}}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{b = \frac{2}{3}}\\{t = 2}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{b = \frac{2}{3}}\\{b = {a^3}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{b = \frac{2}{3}}\\{a = \sqrt[3]{{\frac{2}{3}}}}\end{array}} \right.\)