Ta có \(VT=\dfrac{1}{a}+\dfrac{1}{4b}\)

\(=\dfrac{1}{a}+\dfrac{\dfrac{1}{4}}{b}\)

\(=\dfrac{1^2}{a}+\dfrac{\left(\dfrac{1}{2}\right)^2}{b}\)

\(\ge\dfrac{\left(1+\dfrac{1}{2}\right)^2}{a+b}\) (áp dụng BĐT \(\dfrac{x^2}{m}+\dfrac{y^2}{n}\ge\dfrac{\left(x+y\right)^2}{m+n}\))

\(=\dfrac{\left(\dfrac{3}{2}\right)^2}{1}\) (vì \(a+b=1\))

\(=\dfrac{9}{4}\)

Ta có đpcm. Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\\dfrac{1}{a}=\dfrac{1}{2b}\end{matrix}\right.\) \(\Leftrightarrow\left(a,b\right)=\left(\dfrac{2}{3},\dfrac{1}{3}\right)\)