\(P=4\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\dfrac{1}{4ab}+4ab+\dfrac{11}{4ab}\)
\(P\ge\dfrac{16}{a^2+b^2+2ab}+2\sqrt{\dfrac{4ab}{4ab}}+\dfrac{11}{\left(a+b\right)^2}=\dfrac{27}{\left(a+b\right)^2}+2\ge\dfrac{27}{1}+2=29\)
Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)