Tìm Max của \(\frac{1}{a^3+ab+b^3}+\frac{4a^2b^2+1}{ab}\)
\(\frac{1}{a^3+ab+b^3}+\frac{4a^2b^2+1}{ab}=\frac{1}{\left(a+b\right)\left(a^2-ab+b^2\right)+ab}+\frac{4a^2b^2+1}{ab}\)
\(=\frac{1}{\left(a^2-ab+b^2\right)+ab}+\frac{4a^2b^2+1}{ab}=\frac{1}{\left(a^2+b^2\right)}+\frac{4a^2b^2+1}{ab}\)
\(=\frac{1}{\left(a+b\right)^2-2ab}+\frac{4a^2b^2+1}{ab}=\frac{1}{1-2ab}+4ab+\frac{1}{ab}\)
\(=\left(\frac{1}{1-2ab}+\frac{1}{2ab}\right)+\left(4ab+\frac{1}{4ab}\right)+\frac{1}{4ab}\)
\(\ge\frac{\left(1+1\right)^2}{1-2ab+2ab}+2\sqrt{4ab.\frac{1}{4ab}}+\frac{1}{\left(a+b\right)^2}\)
\(=4+2+1=7\)
Dấu = xảy ra khi \(a=b=\frac{1}{2}\)