Ta có \(A=\left(a+b\right)^2\left(\frac{1}{a^2+b^2}+\frac{1}{ab}\right)\)
\(=\left(a+b\right)^2\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}\right)\)
Áp dụng bđt AM-GM ta có
\(A\ge\left(a+b\right)^2\left(\frac{4}{\left(a+b\right)^2}+\frac{1}{2ab}\right)\)\(=4+\frac{\left(a+b\right)^2}{2ab}\ge4+\frac{4ab}{2ab}=4+2=6\)
Dấu "=" xảy ra khi a=b
Vậy...........