A Basic Product

$\color{#3D99F6}{\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=\frac{(x+1)(y+1)(z+1)}{xyz}}$ We all know that a fraction is smallest when the denominator(in this case $\color{forestgreen}{xyz}$ ) is the largest possible value.For this let,s use the AM-GM inequality: $\color{#69047E}{\frac{x+y+z}{3}\leq\sqrt[3]{xyz}}$ $\color{#3D99F6}{\frac{1}{3}\leq\sqrt[3]{xyz}}$ $\color{#EC7300}{\frac{1}{27}\leq xyz}$ Equality can only occur when $x,y,z=\frac{1}{3}$ .Plugging in this value into the equation,we get: $\color{#3D99F6}{(1+\frac{1}{\frac{1}{3}})^3=(1+3)^3=4^3=\boxed{64}}$

use Am-GM inequality

x = y = z = 1/3 done it