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 inequality between harmonic and geometric mean we have:
$$
\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)\ge\left(\frac{3}{\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}}\right)^3
$$
Now if we prove that 
$$
\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\le\frac{3}{4}
$$
we are done. But it is the same to prove that
$$
\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{3}{\frac{x+1+y+1+z+1}{3}}=\frac{9}{4}.
$$
The last inequality is the inequality between harmonic and arithmetic mean.
