Đáp án:
$MinP = 9 \Leftrightarrow a = b = \dfrac{1}{2}$
Giải thích các bước giải:
ĐK: $a,b>0$
Ta có:
$\begin{array}{l}
P = \left( {1 - \dfrac{1}{{{a^2}}}} \right)\left( {1 - \dfrac{1}{{{b^2}}}} \right)\\
= \dfrac{{\left( {{a^2} - 1} \right)\left( {{b^2} - 1} \right)}}{{{a^2}{b^2}}}\\
= \dfrac{{\left( {1 - {a^2}} \right)\left( {1 - {b^2}} \right)}}{{{a^2}{b^2}}}\\
= \dfrac{{\left( {{{\left( {a + b} \right)}^2} - {a^2}} \right)\left( {{{\left( {a + b} \right)}^2} - {b^2}} \right)}}{{{a^2}{b^2}}}\\
= \dfrac{{\left( {2ab + {b^2}} \right)\left( {2ab + {a^2}} \right)}}{{{a^2}{b^2}}}\\
= \dfrac{{b\left( {2a + b} \right)a\left( {2b + a} \right)}}{{{a^2}{b^2}}}\\
= \dfrac{{\left( {2a + b} \right)\left( {2b + a} \right)}}{{ab}}\\
= \dfrac{{2{a^2} + 5ab + 2{b^2}}}{{ab}}\\
= \dfrac{{2a}}{b} + 5 + \dfrac{{2b}}{a}\\
= 5 + 2\left( {\dfrac{a}{b} + \dfrac{b}{a}} \right)\\
\ge 5 + 2.2\sqrt {\dfrac{a}{b}.\dfrac{b}{a}} \left( {BDT:Cauchy} \right)\\
= 9
\end{array}$
Dấu bằng xảy ra
$ \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{a}{b} = \dfrac{b}{a}\\
a + b = 1
\end{array} \right. \Leftrightarrow a = b = \dfrac{1}{2}$
Vậy $MinP = 9 \Leftrightarrow a = b = \dfrac{1}{2}$
