$ f(x) = \dfrac{x}{2} + \dfrac{2}{x-1} = \dfrac{x-1}{2} + \dfrac{1}{2} + \dfrac{2}{x-1} = ( \dfrac{x-1}{2} + \dfrac{2}{x-1} ) + \dfrac{1}{2} $
Với $ x>1 \to \dfrac{x-1}{2} > 0 ;\ \dfrac{2}{x-1} >0$, áp dụng BĐT Cauchy ta có
$ \dfrac{x-1}{2} + \dfrac{2}{x-1} \ge 2 \sqrt { \dfrac{x-1}{2} . \dfrac{2}{x-1}} = 2$
$\to f(x) \ge 2 + \dfrac{1}{2} = \dfrac{5}{2}$
Vậy $min_{f(x)} = \dfrac{5}{2}$
Dấu $=$ xảy ra khi $ \dfrac{x-1}{2} = \dfrac{2}{x-1} \to (x-1)^2 = 4 \to x = 3$ (tm) hoặc $ x = -1$ (ktm)
Vậy $min_{f(x)} = \dfrac{5}{2}$ khi $ x =3$
