The Other AM-GM (Part 2)

We can see that $\dfrac{1}{w} + \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 5.$

Applying $GM-HM$ on $4w, \space 4w, \space 4w, \space 4w, \space 3x, \space 3x, \space 3x, \space 2y, \space 2y, \space z$ , we have

$\sqrt[10]{256 \cdot 27 \cdot 4 \cdot w^{4}x^{3} y^{2}z} \ge \dfrac{10}{ \underbrace{ \dfrac{1}{4w} + \dfrac{1}{4w} + \dfrac{1}{4w}+ \dfrac{1}{4w} + \dfrac{1}{3x} + \dfrac{1}{3x} +\dfrac{1}{3x} + \dfrac{1}{2y} + \dfrac{1}{2y} + \dfrac{1}{z} }_{ \dfrac{1}{w} + \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{y} + \dfrac{1}{z} = 5 }}$

$\implies \sqrt[10]{2^{10} \cdot 27 w^{4}y^{3}x^{2}z} \ge \dfrac{10}{5} \rightarrow = 2$

$\implies \cancel{2^{10}} \cdot 27 \cdot w^{4}x^{3}y^{2} z \ge \cancel{2^{10}}$

$w^{4}x^{3}y^{2}z \ge \dfrac{1}{27}$

We have $a = 1$ and $b = 27$ .

$\sqrt[3]{\dfrac{a}{b}} = \dfrac{1}{3} = \dfrac{c}{d}$

We have $c = 1$ and $d = 3$ .

Hence, $b - a + d - c = 27 -1 + 3 -1 = \boxed{28}$