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 using Jensen's inequality with the convex function $f(x) = \log(1+\frac1x)$, we get
$$f(x)+f(y)+f(y) \ge 3f\left(\frac{x+y+z}3\right) = 3\log4$$
which on exponentiation gives the inequality.
Use the first equation to rewrite $z$in terms of $x$ and $y$, and then consider
Consider the function $\mathbb R^2\to\mathbb R$:
$$
f(x,y)= \left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{\text{your expression for $z$}}\right)
$$
you are then looking for it's minimum which should be findable by elementary analytic methods.