The least value of the function $f(x)=a x+\dfrac{b}{x}, \quad(x>0, b>0, a>0)$ is

The function is given by $f(x) = ax + \dfrac{b}{x}$ for $x > 0$, $a > 0$, and $b > 0$.

Since $ax > 0$ and $\dfrac{b}{x} > 0$, we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality, which states that for positive real numbers $u$ and $v$, $\dfrac{u+v}{2} \ge \sqrt{uv}$.

Let $u = ax$ and $v = \dfrac{b}{x}$. Then,

$\dfrac{ax + \dfrac{b}{x}}{2} \ge \sqrt{ax \cdot \dfrac{b}{x}}$

$\dfrac{f(x)}{2} \ge \sqrt{ab}$

$f(x) \ge 2\sqrt{ab}$

The equality holds when $ax = \dfrac{b}{x}$, which implies $x^2 = \dfrac{b}{a}$, or $x = \sqrt{\dfrac{b}{a}}$. Since $x > 0$, this value is valid. Thus, the least value of the function is $2\sqrt{ab}$.

Answer: $2 \sqrt{a b}$