Minimum of...

Note that $a=(a-b)+b\ge 2\sqrt{(a-b)b}$ , which implies that $\frac{1}{(a-b)b} \ge \frac{4}{a^2}$ .

Now $\sqrt{2}a^3+\frac{3}{ab-b^2}\ge \sqrt{2}a^3+\frac{12}{a^2}=$ $\frac{\sqrt{2}}{2}a^3+\frac{\sqrt{2}}{2}a^3+\frac{4}{a^2}+\frac{4}{a^2}+\frac{4}{a^2}\ge 5\left(2^5\right)^{\frac{1}{5}}=10$ , where the equality holds iff $a=\sqrt{2}$ and $b=\frac{\sqrt{2}}{2}$ .

Since $a>b>0$ , we can let $a=b+x$ , where $x>0$ . Let $f(a,b)$ be the expression. Then

$\begin{aligned} f(a,b) & = \sqrt 2 a^3 + \frac 3{ab-b^2} \\ & = \sqrt 2 (\blue{b+x})^3 + \frac 3{bx} & \small \blue{\text{By AM-GM inequality}} \\ & \ge \sqrt 2 (\blue{2\sqrt{bx}})^3 + \frac 3{bx} \\ & = 4(\sqrt{2bx})^3 + \frac 6{(\sqrt{2bx})^2} \end{aligned}$

Let $u = \sqrt{2bx}$ and $g(u) = 4u^3 + \dfrac 6{u^2} \implies \dfrac {dg(u)}{du} = 12u^2 - \dfrac {12}{u^3} \implies u = 1$ , when $\dfrac {dg(u)}{du} = 0$ . Since $\dfrac {d^2 g(u)}{du} \bigg|_{u=1} > 0$ , $\min (g(u)) = g(1) = 10$ . Therefore $f(a,b) \ge \boxed{10}$ and equality occurs when $b = x = \dfrac 1{\sqrt 2}$ and $a=\sqrt 2$ .

minimizing $\sqrt{2} a^{3} + \frac{3}{(b)(a-b)}$

-> $\sqrt{2} a^{3} + \frac{3a}{(a)(b)(a-b)}$

-> $\sqrt{2} a^{3} + \frac{3}{a} (\frac{1}{a-b} + \frac{1}{b})$

by AM HM inequality, $\frac{1}{a-b} + \frac{1}{b} >= \frac{4}{a}$

therefore, we have to minimize $\sqrt{2} a^{3} + \frac{12}{a^{2}}.$

break it into $\frac {\sqrt{2}}{2} a^{3} + \frac {\sqrt{2}}{2} a^{3} + \frac{4}{a^{2}} + \frac{4}{a^{2}} + \frac{4}{a^{2}}.$

Applying AM GM, $\frac {\sqrt{2}}{2} a^{3} + \frac {\sqrt{2}}{2} a^{3} + \frac{4}{a^{2}} + \frac{4}{a^{2}} + \frac{4}{a^{2}} >= 5 \times ((\frac {\sqrt{2}}{2})^{2} 64)^{1/5}$ Which gives \( \boxed{10}

\)