Cho các số thực a, b > 1 thỏa mãn giá trị của biểu thức là

* Hướng dẫn giải

Ta có \({a^{{{\log }_b}a}} + 16{b^{{{\log }_a}\left( {\frac{{{b^8}}}{{{a^3}}}} \right)}} = 12{b^2} \Leftrightarrow {a^{{{\log }_b}a}} + 16{b^{\left( {{{\log }_a}{b^8} - {{\log }_a}{a^3}} \right)}} = 12{b^2}\)

\(\begin{array}{l} \Leftrightarrow {a^{{{\log }_b}a}} + 16{b^{\left( {{{\log }_a}{b^8} - {{\log }_a}{a^3}} \right)}} = 12{b^2} \Leftrightarrow {a^{{{\log }_b}a}} + 16{b^{\left( {8{{\log }_a}b - 3} \right)}} = 12{b^2}\\ \Leftrightarrow {a^{{{\log }_b}a}} + 16{b^{\left( {\frac{8}{{{{\log }_b}a}} - 3} \right)}} = 12{b^2}\,\,(*) \end{array}\)

Đặt \({\log _a}b = t \Rightarrow a = {b^t}\). Lại có vì \(a,b > 1 \Rightarrow {\log _a}b > 0\) hay t > 0.

Khi đó ta có

\(VT\left( * \right) = {a^{{{\log }_b}a}} + 16{b^{\left( {\frac{8}{{{{\log }_b}a}} - 3} \right)}} = {\left( {{b^t}} \right)^t} + 16.{b^{\frac{8}{t} - 3}} = {b^{{t^2}}} + 8.{b^{\frac{8}{t} - 3}} + 8.{b^{\frac{8}{t} - 3}}\)  

\(\begin{array}{l} \mathop \ge \limits^{Cô - si} 3\sqrt[3]{{{b^{{t^2}}}.8.{b^{\frac{8}{t} - 3}}8.{b^{\frac{8}{t} - 3}}}} = 12\sqrt[3]{{{b^{{t^2}}}{b^{\frac{8}{t} - 3}}{b^{\frac{8}{t} - 3}}}}12\sqrt[3]{{{b^{{t^2} + \frac{8}{t} + \frac{8}{t} - 6}}}}\\ \mathop \ge \limits^{Cô - si} 12\sqrt[3]{{{b^{3\sqrt[3]{{{t^2}.\frac{8}{t}.\frac{8}{t} - 6}}}}}} = 12\sqrt[3]{{{b^6}}} = 12{b^2}\left( {v\`i \,\,{t^2} + \frac{8}{t} + \frac{8}{t} \ge 3\sqrt[3]{{{t^2}.\frac{8}{t}.\frac{8}{t}}} = 3} \right) \end{array}\)

Hay \(VT\left( * \right) \ge 12{b^2}\), dấu = xảy ra khi

\(\left\{ \begin{array}{l} {b^{{t^2}}} = 8{b^{\frac{8}{t} - 3}}\\ {t^2} = \frac{8}{t} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} t = 2\\ {b^4} = 8b \end{array} \right. \Rightarrow \left\{ \begin{array}{l} t = 2\\ b = 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {\log _b}a = 2\\ b = 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} b = 2\\ a = 4 \end{array} \right.\left( {TM} \right)\)

Suy ra \(P = {a^3} + {b^3} = 64 + 8 = 72\).