Áp dụng BĐT AM-GM ta có:
\(ab\le\dfrac{\left(a+b\right)^2}{4}=\dfrac{1}{4}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{2}{2ab}\)
\(\ge\dfrac{\left(1+1\right)^2}{a^2+b^2+2ab}+\dfrac{2}{2ab}\)
\(\ge\dfrac{4}{\left(a+b\right)^2}+\dfrac{2}{2\cdot\dfrac{1}{4}}=4+\dfrac{2}{\dfrac{1}{2}}=8\)
Xảy ra khi \(a=b=\dfrac{1}{2}\)