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.

Write the means
$$
\begin{align}
H &= \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \\
G &= \sqrt[3]{xyz} \\
A &=\left(\frac{x+y+z}{3}\right)
\end{align}
$$
then the power mean (AM-GM-HM) inequality gives $H \le G \le A$. Also note
$$
xy+yz+zx = xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3G^3}{H}
$$
Now given
$$
\begin{align}
8^3 &= (x^2-1)(y^2-1)(z^2-1) \\
&= (x-1)(y-1)(z-1)(x+1)(y+1)(z+1) \\
&= (xyz-(xy+yz+zx)+(x+y+z)-1)(xyz+(xy+yz+zx)+(x+y+z)+1) \\
&= \left(G^3\left(1-\frac{3}{H}\right)+3A-1\right)
   \left(G^3\left(1+\frac{3}{H}\right)+3A+1\right)
\end{align}
$$
Then we have either
$$
1-\frac{3}{H} < 0 \\
\implies H < 3 \implies
1< \frac{1}{x}+\frac{1}{y}+\frac{1}{z}
$$
or else
$$
1-\frac{3}{H} \ge 0 \\
\implies G^3(1-3/H) \ge H^3(1-3/H) = H^3-3H^2
$$
and hence
$$
\begin{align}
8^3 &= \left(G^3\left(1-\frac{3}{H}\right)+3A-1\right)
   \left(G^3\left(1+\frac{3}{H}\right)+3A+1\right) \\
&\ge (H^3-3H^2+3H-1)(H^3+3H^2+3H+1) \\
&= (H-1)^3(H+1)^3 \\
&= (H^2-1)^3 \\
\implies H &\le 3 \\
\implies 1 &\le \frac{1}{x}+\frac{1}{y}+\frac{1}{z}
\end{align}
$$
