Ta có: \(\left\{\begin{matrix}a^4+b^2\ge2\sqrt{a^4b^2}=2a^2b\\b^4+a^2\ge2\sqrt{b^4a^2}=2b^2a\end{matrix}\right.\)
Do đó \(S\le\frac{1}{2a^2b+2ab^2}+\frac{1}{2b^2a+2a^2b}\)\(=\frac{1}{a^2b+ab^2}\)
\(\le\frac{1}{4ab}\cdot\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{2ab}\le\frac{1}{8}\left(\frac{1}{a}+\frac{1}{b}\right)^2=\frac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=1\)