Does $x,y,z>0$ and $x+y+z=1$ imply $\left(1+\frac 1x\right)\left(1+\frac 1y \right)\left(1+\frac 1z \right)\ge 64$? If $x,y,z$ are positive real numbers such that $x+y+z=1$ then is it true that 
$\left(1+\dfrac 1x\right)\left(1+\dfrac 1y \right)\left(1+\dfrac 1z \right)\ge 64$ ?

By AM-GM, we have 
$$1+x=\frac{1}3+\frac13+\frac13+x\ge 4\sqrt[4]{\frac{x}{3^3}}.$$
So
$$(1+x)(1+y)(1+z)\ge 4^3\sqrt[4]{\frac{xyz}{3^9}}.$$
We then need only show that
$$\sqrt[4]{\frac{xyz}{3^9}}\ge xyz,$$
or that
$$1\ge 3^3xyz.$$
The last inequality is clear thanks to AM-GM.
