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 
$\left(1+\dfrac 1x\right)\left(1+\dfrac 1y \right)\left(1+\dfrac 1z \right)\ge 64$ ?
 A: 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.
A: 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.
A: 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.
A: It is equivalent to show
$$
(x+x+y+z)(x+y+y+z)(x+y+z+z)\geq 64 xyz.
$$
This is just AM-GM.
Use AM-GM to $(x+x+y+z)$, $(x+y+y+z)$, $(x+y+z+z)$.
A: 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.
