\(A=\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}\)
\(=\left(\dfrac{1}{2ab}+\dfrac{1}{a^2+b^2}\right)+\dfrac{1}{2ab}\)
\(\ge\dfrac{\left(1+1\right)^2}{a^2+2ab+b^2}+\dfrac{1}{2.\dfrac{\left(a+b\right)^2}{4}}\)
\(=\dfrac{4}{\left(a+b\right)^2}+\dfrac{1}{\dfrac{2.1}{4}}=6\)
Đẳng thức xảy ra \(\Leftrightarrow a=b=\dfrac{1}{2}\)