Inequality $2a^2 + 8b^2 + \frac{1}{2ab} \ge 4$ I saw the following question:
Prove that the least value of
$$2a^2+8b^2+\frac{1}{2ab}$$
is 4, where $a,b$ are positive numbers.
I believe that this can be shown using the criterion of the saddle and critical points that is taught usually in Calculus 3.
Can this be solved using another easier method?

Differently from the proposed solutions, you can also proceed as follows:
\begin{align*}
2a^{2} + 8b^{2} + \frac{1}{2ab} & = 2(a^{2} + 4b^{2}) + \frac{1}{2ab}\\\\
& = 2(a^{2} - 4ab + 4b^{2}) + 8ab + \frac{1}{2ab}\\\\
& = 2(a - 2b)^{2} + \frac{1}{2ab} + 8ab\\\\
& \geq \frac{1}{2ab} + 8ab
\end{align*}
Finally, according to the AM-GM inequality, one arrives at
\begin{align*}
2a^{2} + 8b^{2} + \frac{1}{2ab} \geq \frac{1}{2ab} + 8ab \geq 2\sqrt{\frac{8ab}{2ab}} = 4
\end{align*}
and we are done.
Hopefully this helps !
