\(b\left(a-b\right)\le\dfrac{\left(b+a-b\right)^2}{4}=\dfrac{a^2}{4}\)
\(\Rightarrow\dfrac{1}{b\left(a-b\right)}\ge\dfrac{4}{a^2}\)
\(\Rightarrow a+\dfrac{1}{b\left(a-b\right)}\ge a+\dfrac{4}{a^2}=\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{4}{a^2}\ge3\sqrt[3]{\dfrac{a}{2}\dfrac{a}{2}\dfrac{4}{a^2}}=3\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{a}{2}=\dfrac{4}{a^2}\\b=a-b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
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