Inequality

$\dfrac{a+bx^4}{x^2}\\ =bx^2 + \dfrac{a}{x^2}\\ =bx^2 + 2\left(\sqrt{b}x\right)\left(\dfrac{\sqrt{a}}{x}\right) - 2\left(\sqrt{b}x\right)\left(\dfrac{\sqrt{a}}{x}\right) + \dfrac{a}{x^2}\\ =\left(\sqrt{b}x\right)^2 - 2\sqrt{ab} + \left(\dfrac{\sqrt{a}}{x}\right)^2 + 2\sqrt{ab}\\ =\left(\sqrt{b}x - \dfrac{\sqrt{a}}{x}\right)^2 + 2\sqrt{ab}$

The minimum value of this expression occurs when

$\sqrt{b}x - \dfrac{\sqrt{a}}{x} = 0\\ \sqrt{b}x = \dfrac{\sqrt{a}}{x}\\ x^2 = \sqrt{\dfrac{a}{b}}$

Therefore, $k=\boxed{1}$

$\frac{a+bx^4}{x^2}=\frac{a}{x^2}+bx^2\stackrel{\text{AM-GM}}\geq 2\sqrt{\frac{a}{x^2}\times bx^2}\\ \frac{a}{x^2}+bx^2\geq 2\sqrt{ab}$ Equality occurs when $\dfrac{a}{x^2}=bx^2\implies x^4=\dfrac{a}{b}\implies x^2=\sqrt{\dfrac{a}{b}}\implies \boxed{k=1}$

If we substitute $x^2$ and $x^4$ by $k\sqrt{\frac{a}{b}}$ and $k^2\frac{a}{b}$ respectively, we get:

$\frac{a+bk^2\frac{a}{b}}{k\sqrt{\frac{a}{b}}} = \frac{\sqrt{ab}}{k} + \sqrt{ab}k = \sqrt{ab}(k + \frac{1}{k})$

The minimum value of this expression is when k = 1.

Using AM-GM, we get $a+bx^4\geq 2x^2\sqrt{ab} \Rightarrow \frac{a+bx^4}{x^2}\geq 2\sqrt{ab}$ . The equality happens when $x^4=\frac{a}{b}\Rightarrow x^2=k\sqrt{\frac{a}{b}}=\sqrt{\frac{a}{b}}\therefore k=1$

Let $f(x) = \dfrac{a+bx^4}{x^2}$

$f(x) = \dfrac{a}{x^2} + {bx^2}$

$f'(x) = \dfrac{-2a}{x^3} +2bx$

$f''(x) = \dfrac{6a}{x^4} +2b > 0$

$f(x)$ has minimum value. $f'(x) = \dfrac{-2a}{x^3} +2bx = 0 \implies x^2 = \sqrt{\frac{a}{b}}$

therefore $k= 1$