Sửa đề chút:
\(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge\frac{25}{2}\)
Áp dụng BĐT Cauchy-schwarz ta có: ( link chứng minh: Xem câu hỏi)
\(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge\frac{\left(a+\frac{1}{a}+b+\frac{1}{b}\right)^2}{1+1}=\frac{\left[1+\frac{\left(1+1\right)^2}{a+b}\right]^2}{2}=\frac{\left(4+1\right)^2}{2}=\frac{25}{2}\)Dấu " = " xảy ra \(\Leftrightarrow a=b=\frac{1}{2}\)
\(ab\le\frac{\left(a+b\right)^2}{4}=\frac{1}{4}\Rightarrow\frac{1}{ab}\ge4\)
\(P=\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2=a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}+4\)
\(P\ge2ab+\frac{2}{ab}+4=2ab+\frac{1}{8ab}+\frac{15}{8ab}+4\)
\(P\ge2\sqrt{\frac{2ab}{8ab}}+\frac{15}{8}.4+4=\frac{25}{2}\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
