- Sửa đề: \(\left\{{}\begin{matrix}a,b>0\\a+b\le1\end{matrix}\right.\)
\(S=\dfrac{1}{a^3+b^3}+\dfrac{1}{a^2b}+\dfrac{1}{ab^2}\)
\(=\dfrac{1}{\left(a+b\right)\left(a^2-ab+b^2\right)}+\dfrac{1}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(\ge\dfrac{1}{a^2-ab+b^2}+\dfrac{1}{ab}.\dfrac{4}{a+b}\)
\(\ge\dfrac{1}{a^2-ab+b^2}+\dfrac{1}{ab}.\dfrac{4}{1}\)
\(=\dfrac{1}{a^2-ab+b^2}+\dfrac{4}{ab}\)
\(=\dfrac{1}{a^2-ab+b^2}+\dfrac{1}{ab}+\dfrac{3}{ab}\)
\(\ge\dfrac{\left(1+1\right)^2}{a^2-ab+b^2+ab}+\dfrac{3}{ab}\)
\(=\dfrac{4}{a^2+b^2}+\dfrac{3}{ab}\)
\(=\dfrac{4}{a^2+b^2}+\dfrac{4}{2ab}+\dfrac{1}{ab}\)
\(\ge\dfrac{\left(2+2\right)^2}{a^2+b^2+2ab}+\dfrac{1}{\dfrac{\left(a+b\right)^2}{4}}\)
\(\ge\dfrac{16}{\left(a+b\right)^2}+\dfrac{1}{\dfrac{1}{4}}\)
\(\ge\dfrac{16}{1^2}+4=20\)
- Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)
- Vậy \(MinS=20\)
