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 $a = kb, k > 1$.
Denote $c = a - b > 0$, then$$ \frac{a + b}{2} + \frac{2}{ab - b^2} = b + \frac{c}{2} + \frac{2}{bc} \geqslant 3 \sqrt[3]{b · \frac{c}{2} · \frac{2}{bc}} = 3. $$ The equality is achieved at $b = 1$, $c = 2$, i.e. $(a, b) = (3, 1)$.