Prove $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 1$ when $(x^2-1)(y^2-1)(z^2-1)=8^3$ Please help to prove this inequality.
Prove $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 1$ when $(x^2-1)(y^2-1)(z^2-1)=8^3$ and each of $x,y,z$ is greater than 1.
Thanks.

By symmetry, the minimum value of the expression
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
under the constraint $(x^2-1)(y^2-1)(z^2-1)=8^3$ will occur when $x=y=z$.  From this, it follows that the minimum occurs when $x=y=z=3$, and the minimum is $1$.
