Ta có:
\(a+b\ge2\sqrt{ab}\)
\(\Rightarrow1\ge2\sqrt{ab}\)
\(\Leftrightarrow ab\le\frac{1}{4}\)
Quay lại bài toán ta có:
\(K=\frac{1}{ab}+\frac{1}{a^2+b^2}=\frac{1}{2ab}+\left(\frac{1}{2ab}+\frac{1}{a^2+b^2}\right)\)
\(\ge\frac{1}{\frac{2}{4}}+\frac{4}{\left(a+b\right)^2}=2+4=6\)
Dấu = xảy ra khi \(a=b=\frac{1}{2}\)