parmenides51 wrote: Let $a>b>0$. Prove that $\sqrt2 a^3+ \frac{3}{ab-b^2}\ge 10$ When does equality hold? $$\sqrt2 a^3+ \frac{3}{ab-b^2}\ge3(a^2+\frac{4}{a^2})-2\geq 10.$$Equality holds when $a=2b=\sqrt 2.$

Can you elaborate on your solution, I, unfortunately, don't get why that is true.

Yes, can you please elaborate on your solution. I understand that you used AM-GM in the second part but how did you get from the first expression to the second.

$$\sqrt{2}a^3+2\geq 3a^2, ab-b^2\leq\left(\frac{b+a-b}{2}\right)^2=\frac{a^2}{4}$$ Let $a>b>0$. Prove that $$81a^3+\frac{4}{ab-b^2}\geq 60$$here

By AMGM \begin{align*} b(a - b) &\leq \frac {a^2}{4} \\ \sqrt{2}a^3 + \frac{12}{a^2} &\geq 10 \rightarrow 2 \frac{\sqrt{2} a^3}{2} + 3 \frac{4}{a^2} \geq 10 \\ \text{Equality holds if } a &= 2b = \sqrt2 \end{align*}