Cho các số thực \(a, b >1\) thỏa mãn \({a^{{{\l...

D

- Hướng dẫn giải

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

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

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

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

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

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