Áp dụng BĐT AM-GM ta có:
\(A=a+b+\dfrac{1}{a}+\dfrac{1}{b}\)
\(=\left(a+\dfrac{1}{4a}\right)+\left(b+\dfrac{1}{4b}\right)+3\left(\dfrac{1}{4a}+\dfrac{1}{4b}\right)\)
\(=2\sqrt{a\cdot\dfrac{1}{4a}}+2\sqrt{b\cdot\dfrac{1}{4b}}+3\dfrac{\left(1+1\right)^2}{4\left(a+b\right)}\)
\(\ge2\cdot\dfrac{1}{2}+2\cdot\dfrac{1}{2}+\dfrac{3\cdot4}{4}=5=VP\)
Xảy ra khi \(a=b=\dfrac{1}{2}\)