<s>[INST] 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?
 [/INST] 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 !
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