Đáp án: 4

Giải chi tiết:

ĐKXĐ: \(\left\{ {\begin{array}{*{20}{l}}{b - a > 0}\\{a,{\mkern 1mu} {\mkern 1mu} b > 0}\end{array}} \right.\).

Ta có: \({\log _3}\left( {1 + ab} \right) = \frac{1}{2} + {\log _3}\left( {b - a} \right)\)

\( \Leftrightarrow {\log _3}\left( {1 + ab} \right) - {\log _3}\left( {b - a} \right) = \frac{1}{2}\)

\( \Leftrightarrow {\log _3}\frac = \frac{1}{2}\)

\( \Leftrightarrow \frac = \sqrt 3 \)

\( \Leftrightarrow 1 + ab = \sqrt 3 \left( {b - a} \right)\)

\( \Leftrightarrow \frac{1}{a} + b = \sqrt 3 \left( {\frac{b}{a} - 1} \right)\).

Áp dụng BĐT Cô-si ta có \(\frac{1}{a} + b \ge 2\sqrt {\frac{b}{a}} \) nên

\(\sqrt 3 \left( {\frac{b}{a} - 1} \right) \ge 2\sqrt {\frac{b}{a}} \Leftrightarrow \sqrt 3 \frac{b}{a} - 2\sqrt {\frac{b}{a}} - \sqrt 3 \ge 0\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\sqrt {\frac{b}{a}} \ge \sqrt 3 }\\{\sqrt {\frac{b}{a}} \le - \frac{1}{{\sqrt 3 }}{\mkern 1mu} {\mkern 1mu} \left( {Loai} \right)}\end{array}} \right. \Leftrightarrow \sqrt {\frac{b}{a}} \ge \sqrt 3 \Leftrightarrow \frac{b}{a} \ge 3\)

Ta có: \(P = \frac{{\left( {1 + {a^2}} \right)\left( {1 + {b^2}} \right)}}{{a\left( {a + b} \right)}} = \frac}{{a\left( {a + b} \right)}}\)

Áp dụng BĐT Cô-si ta có \(1 + {a^2}{b^2} \ge 2\sqrt {{a^2}{b^2}} = 2ab\) nên

\(1 + {a^2} + {b^2} + {a^2}{b^2} \ge {a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}\)

\( \Rightarrow P = \frac}{{a\left( {a + b} \right)}} \ge \frac{{{{\left( {a + b} \right)}^2}}}{{a\left( {a + b} \right)}} = \frac{a} = 1 + \frac{b}{a} \ge 4\)

Vậy \({P_{\min }} = 4 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{\frac{1}{a} = b}\\{\frac{b}{a} = 3}\\{a,{\mkern 1mu} {\mkern 1mu} b > 0,{\mkern 1mu} {\mkern 1mu} b - a > 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{\frac{1}{a} = 3a}\\{b = 3a}\\{a,{\mkern 1mu} {\mkern 1mu} b > 0,{\mkern 1mu} {\mkern 1mu} b - a > 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{a = \frac{1}{{\sqrt 3 }}}\\{b = \sqrt 3 }\end{array}} \right.\).