Đáp án:
\(\min Q = 22 \Leftrightarrow a = b = \dfrac12\)
Giải thích các bước giải:
\(\begin{array}{l}
\text{Ta có:}\\
\quad a + b \geqslant 2\sqrt{ab}\quad (BDT\ Cauchy)\\
\to (a+b)^2 \geqslant 4ab\\
\to 1 \geqslant 4ab\\
\to \dfrac{1}{ab} \geqslant 4\\
\text{Ta được:}\\
\quad Q = \dfrac{1}{a^2 + b^2} + \dfrac{5}{ab}\\
\to Q = \dfrac{1}{a^2 + b^2} + \dfrac{1}{2ab} + \dfrac{9}{2ab}\\
\to Q \geqslant \dfrac{(1+1)^2}{a^2 + 2ab + b^2} + \dfrac{9}{2ab}\quad (BDT\ Cauchy-Schwarz)\\
\to Q \geqslant \dfrac{4}{(a+b)^2} + \dfrac{9}{2}\cdot \dfrac{1}{ab}\\
\to Q \geqslant \dfrac{4}{1^2} + \dfrac{9}{2}\cdot 4\\
\to Q \geqslant 22\\
\text{Dấu = xảy ra}\ \Leftrightarrow \begin{cases}a^2 + b^2 = 2ab\\a + b = 1\\a = b\end{cases}\Leftrightarrow a = b= \dfrac12\\
\text{Vậy}\ \min Q = 22 \Leftrightarrow a = b = \dfrac12
\end{array}\)
