Correct Option: (C)Explanation:We are asked to find the minimum value of the expression $ E = \frac{(x^2+1)(y^2+4)(z^2+9)}{xyz} $ where $x, y, z$ are positive real numbers.The expression can be algebraically manipulated by dividing each factor in the numerator by one of the variables in the denominator:$ E = \left(\frac{x^2+1}{x}\right) \left(\frac{y^2+4}{y}\right) \left(\frac{z^2+9}{z}\right) $Further simplifying each term within the parentheses:$ E = \left(x+\frac{1}{x}\right) \left(y+\frac{4}{y}\right) \left(z+\frac{9}{z}\right) $Now, we apply the Arithmetic Mean - Geometric Mean (AM-GM) inequality to each factor. The AM-GM inequality states that for any two positive real numbers $a$ and $b$, $a+b \ge 2\sqrt{ab}$.1. For the first factor, $x+\frac{1}{x}$: Since $x > 0$, we have $x+\frac{1}{x} \ge 2\sqrt{x \cdot \frac{1}{x}} = 2\sqrt{1} = 2$. Equality holds when $x = \frac{1}{x}$, which implies $x^2=1$. Since $x$ is positive, $x=1$.2. For the second factor, $y+\frac{4}{y}$: Since $y > 0$, we have $y+\frac{4}{y} \ge 2\sqrt{y \cdot \frac{4}{y}} = 2\sqrt{4} = 4$. Equality holds when $y = \frac{4}{y}$, which implies $y^2=4$. Since $y$ is positive, $y=2$.3. For the third factor, $z+\frac{9}{z}$: Since $z > 0$, we have $z+\frac{9}{z} \ge 2\sqrt{z \cdot \frac{9}{z}} = 2\sqrt{9} = 6$. Equality holds when $z = \frac{9}{z}$, which implies $z^2=9$. Since $z$ is positive, $z=3$.Since the conditions for equality for each factor ($x=1, y=2, z=3$) can be satisfied simultaneously, the minimum value of the entire product is the product of the individual minimum values of each factor.Minimum value of $E = (2)(4)(6) = 48$.Therefore, the minimum value of the expression is 48.