Find the minimum value of $\frac{a+b}{2} + \frac{2}{ab-b^{2}}$, with $a,b \in \mathbb{R}$, $a>b>0$ Find the minimum value of $$ \frac{a+b}{2} + \frac{2}{ab-b^{2}},$$ 
where $a,b \in \mathbb{R}$, $a>b>0$.

Attempt : The only method I knew was using partial derivatives. 
Let 
$$f(a,b) = \frac{a+b}{2} + \frac{2}{ab-b^{2}},$$
then the partial derivatives are 
$$ f_{a} = \frac{1}{2} - \frac{2b}{(ab-b^{2})^{2}},\\
f_{b} =  \frac{1}{2} - \frac{2(a-2b)}{(ab-b^{2})^{2}}.$$
Setting them to $0$ implies
$$ \frac{2b}{(ab-b^{2})^{2}} = \frac{2(a-2b)}{(ab-b^{2})^{2}},$$
so
$$ 3b = a  $$
with $ ab - b^{2} \ne 0$.
Subtitute this to $f(a,b)$:
$$ f(b) = 2b + \frac{1}{b^{2}} \implies f'(b) = \frac{2b^{3} - 2}{b^{3}}.$$ 
so we must have $b^{3} = 1 \implies b = 1$, for $f'(b)=0$. So this means $a=3$. Then
$$f(a,b) = 3.$$

How to check whether this is valid minimum global of the expression, preferably without testing $D(a,b) = f_{aa} f_{bb} - (f_{ab})^{2}$? 
How to solve another way without multivariable calculus? One way perhaps is by letting $a = kb, k > 1$.

Using A.M. >= G.M.
$$\cfrac{a-b}{2} + {b} + \cfrac{2}{b(a-b)} >=3\sqrt[3]{\cfrac{a-b}{2}*{b}* \cfrac{2}{b(a-b)}} = 3$$
