Giải thích các bước giải:
ĐKXĐ: $a,b\ge 0; a,b\ne 0$
Ta có:
$\left( {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} \right){\left( {a + b} \right)^2} = \dfrac{{{a^2} + {b^2}}}{{{{\left( {ab} \right)}^2}}}{\left( {a + b} \right)^2}$
Áp dụng BĐT Cauchy ta có:
$\left\{ \begin{array}{l}
{a^2} + {b^2} \ge 2\sqrt {{a^2}{b^2}} = 2ab\left( {a,b \ge 0} \right)\\
a + b \ge 2\sqrt {ab}
\end{array} \right.$
Như vậy:
$\begin{array}{l}
\left( {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} \right){\left( {a + b} \right)^2} = \dfrac{{{a^2} + {b^2}}}{{{{\left( {ab} \right)}^2}}}{\left( {a + b} \right)^2} \ge \dfrac{{2ab}}{{{{\left( {ab} \right)}^2}}}.{\left( {2\sqrt {ab} } \right)^2} = 8\\
\Leftrightarrow \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} \ge \dfrac{8}{{{{\left( {a + b} \right)}^2}}}
\end{array}$
Dấu bằng xảy ra $ \Leftrightarrow a = b$
Ta có ĐPCM
