Áp dụng BĐT AM-GM, ta có:
\(\dfrac{1}{ab}\ge\dfrac{4}{\left(a+b\right)^2}\)
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{\left(a+b\right)^2}\)
Ta có:
\(\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}=\dfrac{1}{2}.\dfrac{1}{ab}+\dfrac{1}{2}.\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}\)
\(=\dfrac{1}{2}.\dfrac{1}{ab}+\dfrac{1}{2ab}+\dfrac{1}{a^2+b^2}\)
\(\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}\ge\dfrac{1}{2}.\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(a^2+b^2+2ab\right)^2}\)
\(\ge\dfrac{1}{2}.\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left[\left(a+b\right)^2\right]^2}\)
\(\ge\dfrac{1}{2}.\dfrac{4}{1^2}+\dfrac{4}{\left(1^2\right)^2}\)
\(\ge2+4=6\) ( đfcm )
