Ta có : \(a+b=2\Rightarrow\left(a+b\right)^2=4\)
Theo BĐT Bu-nhi-a-cốp-xki ta có :
\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\Leftrightarrow a^2+b^2\ge2\)
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge\dfrac{4}{a^2+b^2}=2\)
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}\times\dfrac{b}{a}}=2\)
\(\Rightarrow Q\ge2.2-6.2+9.2=10\)
Dấu \("="\) xảy ra khi \(a=b=1\)