AM-GM Inequality Applied

Our function is $f(a,b)=a+\dfrac{1}{b(a-b)}$ . This function attains an optimum when $\dfrac{\partial f}{\partial a}=0, \dfrac{\partial f}{\partial b}=0\implies 1-\dfrac{1}{b(a-b)^2}=0, a-2b=0\implies a=2b, b(a-b)^2=1\implies b^2=1\implies b=1, a=2$ (since $b$ is positive). The optimum value of the function is $2+\dfrac{1}{1(2-1)}=2+1=\boxed 3$ . At these values of $a$ and $b$ , the value of the function is minimum , which can easily be checked from the second derivatives.

Since $a>b$ . let $a=b+c$ , where $c$ is a positive real. Then by AM-GM inequality , we have:

$b+c + \frac 1{bc} \ge 3\sqrt[3]{\frac {bc}{bc}} = \boxed 3$

Equality occurs when $b=c=1$ or $a = 2$ .

By AM-GM inequality, we have $a+\frac{1}{b(a-b)}=b+(a-b)+\frac{1}{b(a-b)}\ge 3.$ Equality happens when $b=1, a=2$ .