Xét các số thực dương $a$ và $b$ thỏa mãn ${{\log }_{3}}\left( 1+ab \right)=\dfrac{1}{2}+{{\log }_{3}}\left( b-a \right).$ Giá trị nhỏ nhất của...

Cách giải:
ĐKXĐ: $\left\{ \begin{aligned}
& b-a>0 \\
& a,b>0 \\
\end{aligned} \right..$
Ta có:
${{\log }_{3}}\left( 1+ab \right)=\dfrac{1}{2}+{{\log }_{3}}\left( b-a \right)$
$\Leftrightarrow {{\log }_{3}}\left( 1+ab \right)-{{\log }_{3}}\left( b-a \right)=\dfrac{1}{2}$
$\Leftrightarrow {{\log }_{3}}\dfrac{1+ab}{b-a}=\dfrac{1}{2}$
$\Leftrightarrow \dfrac{1+ab}{b-a}=\sqrt{3}$
$\Leftrightarrow 1+ab=\sqrt{3}\left( b-a \right)$
$\Leftrightarrow \dfrac{1}{a}+b=\sqrt{3}\left( \dfrac{b}{a}-1 \right)$
Áp dụng BĐT Cô-si ta có $\dfrac{1}{a}+b\ge 2\sqrt{\dfrac{b}{a}}$ nên $\sqrt{2}\left( \dfrac{b}{a}-1 \right)\ge 2\sqrt{\dfrac{b}{a}}\Leftrightarrow \sqrt{3}\dfrac{b}{a}-2\sqrt{\dfrac{b}{a}}-\sqrt{3}\ge 0$
$\Leftrightarrow \left[ \begin{aligned}
& \sqrt{\dfrac{b}{a}}\ge \sqrt{3} \\
& \sqrt{\dfrac{b}{a}}\le -\dfrac{1}{\sqrt{3}}\left( Loai \right) \\
\end{aligned} \right.\Leftrightarrow \sqrt{\dfrac{b}{a}}\ge \sqrt{3}\Leftrightarrow \dfrac{b}{a}\ge 3$
Ta có: $P=\dfrac{\left( 1+{{a}^{2}} \right)\left( 1+{{b}^{2}} \right)}{a\left( a+b \right)}=\dfrac{1+{{a}^{2}}+{{b}^{2}}+{{a}^{2}}{{b}^{2}}}{a\left( a+b \right)}$
Áp dụng BĐT Cô-si ta có $1+{{a}^{2}}{{b}^{2}}\ge 2\sqrt{{{a}^{2}}{{b}^{2}}}=2ab$ nên $1+{{a}^{2}}+{{b}^{2}}+{{a}^{2}}{{b}^{2}}\ge {{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$
$\Rightarrow P=\dfrac{1+{{a}^{2}}+{{b}^{2}}+{{a}^{2}}{{b}^{2}}}{a\left( a+b \right)}\ge \dfrac{{{\left( a+b \right)}^{2}}}{a\left( a+b \right)}=\dfrac{a+b}{a}=1+\dfrac{b}{a}\ge 4.$
Vậy ${{P}_{\min }}=4\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{a}=b \\
& \dfrac{b}{a}=3 \\
& a,b>0,b-a>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{a}=3a \\
& b=3a \\
& a,b>0,b-a>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{1}{\sqrt{3}} \\
& b=\sqrt{3} \\
\end{aligned} \right..$