Finding the minimum value of a function without using Calculus

Use AM-GM: $$x^4+\frac1{x^2}=x^4+\frac1{2x^2}+\frac1{2x^2}\ge 3\sqrt[3]{\frac1{4}},$$ equality occurs when $x^4=\frac1{2x^2}=\frac1{2x^2} \Rightarrow x=\pm\frac1{\sqrt[6]{2}}$.

Hint:

As $x^4,1/x^2>0$ using Weighted Form of Arithmetic Mean-Geometric Mean Inequality

$$\dfrac{ax^4+bx^{-2}}{a+b}\ge\sqrt[a+b]{x^{4a-2b}}$$

Set $4a-2b=0\iff b=2a$

First, the minimum value of $x$ is $b$ such that $f(x) - b$ has a (double) root. (That is, the amount you must shift the graph of $f$ down so that it meets the $x$-axis once.) So we are looking for a $b$ so that $f(x) - b = x^4 + \frac{1}{x^2} - b = 0$ has a (double) root. Observe $$ x^4 + \frac{1}{x^2} - b = \frac{x^6 - b x^2 + 1}{x^2} $$ has a root exactly when its numerator does. So now we just need to know when that cubic in $x^2$ has a double root.

The discriminant of $(x^2)^3 - b(x^2) + 1$ is $$ -4(-b)^3 - 27 \cdot (1)^2 = 4b^3 - 27 \text{.} $$ The discriminant is zero if and only if the polynomial has a double root. Taking $b = \sqrt[3]{27/4} = \frac{3}{2^{2/3}}$ is the only choice that makes the discriminant zero, so the minimum value of $f$ is this $b$.