Đáp án:
\[{Q_{\min }} = 4 \Leftrightarrow a = b = 1\]
Giải thích các bước giải:
Áp dụng bất đẳng thức Bunhia- Copski ta có:
\(\begin{array}{l}
Q = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{2}{{ab}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{ab}} + \frac{1}{{ab}}\\
\Leftrightarrow 4.Q = 4.\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{ab}} + \frac{1}{{ab}}} \right)\\
\Leftrightarrow 4Q = {\left( {a + b} \right)^2}.\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{ab}} + \frac{1}{{ab}}} \right)\\
\Leftrightarrow 4Q = \left( {{a^2} + {b^2} + ab + ab} \right).\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{ab}} + \frac{1}{{ab}}} \right)\\
\Leftrightarrow 4Q \ge {\left( {a.\frac{1}{a} + b.\frac{1}{b} + \sqrt {ab} .\frac{1}{{\sqrt {ab} }} + \sqrt {ab} .\frac{1}{{\sqrt {ab} }}} \right)^2}\\
\Leftrightarrow 4Q \ge {4^2}\\
\Leftrightarrow Q \ge 4
\end{array}\)
Vậy \({Q_{\min }} = 4 \Leftrightarrow a = b = 1\)
