Solution: The correct option is B).Let the given expression be $E = \frac{1}{x^3(y+z)} + \frac{1}{y^3(z+x)} + \frac{1}{z^3(x+y)}$.Given that $x, y, z$ are positive real numbers and $xyz = 1$.To simplify the expression, we can use a substitution. Let $x = \frac{1}{a}$, $y = \frac{1}{b}$, $z = \frac{1}{c}$.Since $xyz = 1$, we have $\frac{1}{a} \cdot \frac{1}{b} \cdot \frac{1}{c} = 1$, which implies $abc = 1$. As $x,y,z > 0$, it follows that $a,b,c > 0$.Substitute these into the expression $E$:$E = \frac{1}{(\frac{1}{a})^3(\frac{1}{b}+\frac{1}{c})} + \frac{1}{(\frac{1}{b})^3(\frac{1}{c}+\frac{1}{a})} + \frac{1}{(\frac{1}{c})^3(\frac{1}{a}+\frac{1}{b})}$$E = \frac{a^3}{\frac{b+c}{bc}} + \frac{b^3}{\frac{c+a}{ca}} + \frac{c^3}{\frac{a+b}{ab}}$$E = \frac{a^3bc}{b+c} + \frac{b^3ca}{c+a} + \frac{c^3ab}{a+b}$Since $abc = 1$, we can replace $bc$ with $\frac{1}{a}$, $ca$ with $\frac{1}{b}$, and $ab$ with $\frac{1}{c}$.$E = \frac{a^3(\frac{1}{a})}{b+c} + \frac{b^3(\frac{1}{b})}{c+a} + \frac{c^3(\frac{1}{c})}{a+b}$$E = \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b}$Now, we apply Titu's Lemma (a specific form of Cauchy-Schwarz inequality), which states that for positive real numbers $x_i$ and $y_i$: $\sum_{i=1}^n \frac{x_i^2}{y_i} \ge \frac{(\sum_{i=1}^n x_i)^2}{\sum_{i=1}^n y_i}$.Applying this to $E$:$E \ge \frac{(a+b+c)^2}{(b+c)+(c+a)+(a+b)}$$E \ge \frac{(a+b+c)^2}{2(a+b+c)}$$E \ge \frac{a+b+c}{2}$To find the minimum value of $E$, we need to find the minimum value of $a+b+c$.By the AM-GM inequality, for positive real numbers $a, b, c$:$\frac{a+b+c}{3} \ge \sqrt[3]{abc}$Since $abc = 1$:$\frac{a+b+c}{3} \ge \sqrt[3]{1}$$\frac{a+b+c}{3} \ge 1$$a+b+c \ge 3$Substituting this minimum value of $a+b+c$ into our inequality for $E$:$E \ge \frac{3}{2}$The minimum value of the expression is $\frac{3}{2}$. This minimum occurs when $a=b=c=1$, which in turn implies $x=y=z=1$.If $x=y=z=1$, then $E = \frac{1}{1^3(1+1)} + \frac{1}{1^3(1+1)} + \frac{1}{1^3(1+1)} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}$.Thus, the minimum value is $\frac{3}{2}$.