Ta có : \(A=\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{4}{2ab}\)
Sử dụng BĐT Bunhiacopxki ta có :
\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{4}{2ab}=\frac{1^2}{a^2}+\frac{1^2}{b^2}+\frac{2^2}{2ab}\ge\frac{\left(1+1+2\right)^2}{a^2+b^2+2ab}\)
\(=\frac{4^2}{\left(a+b\right)^2}=\frac{16}{2^2}=\frac{16}{4}=4\)
Dấu = xảy ra khi và chỉ khi \(a=b=1\)
Vậy \(A_{min}=4\)khi \(a=b=1\)
\(A=\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{4}{2ab}\)
\(\ge\frac{\left(1+1+2\right)^2}{a^2+2ab+b^2}=\frac{16}{\left(a+b\right)^2}=\frac{16}{4}=4\)
Dấu "=" xảy ra <=> a = b = 1
