\(3a+3b+\dfrac{1}{a+b}=\dfrac{a+b}{25}+\dfrac{1}{a+b}+\dfrac{74\left(a+b\right)}{25}\ge2.\sqrt{\dfrac{a+b}{25}.\dfrac{1}{a+b}}+\dfrac{74}{25}.5=\dfrac{76}{5}\)
Dấu "=" xảy ra khi \(a=b=\dfrac{5}{2}\)
Vậy GTNN của biểu thức là \(\dfrac{76}{5}\)
Ta có: 3a + 3b + \(\dfrac{1}{a+b}\) = \(\dfrac{1}{a+b}+\dfrac{a+b}{25}+\dfrac{74}{25}\left(a+b\right)\)
Áp dụng BDT Co-si, ta có:
\(\dfrac{1}{a+b}+\dfrac{a+b}{25}\ge2\sqrt{\dfrac{1}{a+b}.\dfrac{a+b}{25}}\)
=> \(\dfrac{1}{a+b}+\dfrac{a+b}{25}\ge\dfrac{2}{5}\)
Mà \(\dfrac{74}{25}\left(a+b\right)\ge\dfrac{74}{5}\)
=> \(3\left(a+b\right)+\dfrac{1}{a+b}\ge\dfrac{76}{5}\)
Dấu "=" xảy ra <=> \(a=b=\dfrac{5}{2}\)
