Problem: The problem is to find the minimum value of $2x^2+\frac{1}{x^4}$ and though you can use any method in finding the answer, you probably want to use AM-GM method.
My thought process before getting stuck In this case I know that according to the AM-GM theorem/equation $\frac{a_1+a_2+\cdots a_n}{n}\ge\sqrt[n]{a_1a_2\cdots a_n}$. But then I do not now how to use the AM-GM method in finding the minimum value.
On
The equation $$ \frac{a+b+c}2\left[(b-c)^2+(c-a)^2+(a-b)^2\right]=a^3+b^3+c^3-3abc $$ shows that equality in the AGM $$ \sqrt[\large3]{xyz}\le\frac{x+y+z}3 $$ occurs only when $x=y=z$.
This in conjunction with Jason M's observation that the AGM applies to this question as $$ 3\le x^2+x^2+\frac1{x^4} $$ shows that the minimum is $3$ and it is attained when $x=1$.
$2x^2+\frac{1}{x^4}=x^2+x^2+\frac{1}{x^4} \geq 3 \sqrt[3]{(x^2)(x^2)\left(\frac{1}{x^4}\right)}=3$