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
$$
\begin{align}
H &amp;= \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \\
G &amp;= \sqrt[3]{xyz} \\
A &amp;=\left(\frac{x+y+z}{3}\right)
\end{align}
$$
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
$$
\begin{align}
8^3 &amp;= (x^2-1)(y^2-1)(z^2-1) \\
&amp;= (x-1)(y-1)(z-1)(x+1)(y+1)(z+1) \\
&amp;= (xyz-(xy+yz+zx)+(x+y+z)-1)(xyz+(xy+yz+zx)+(x+y+z)+1) \\
&amp;= \left(G^3\left(1-\frac{3}{H}\right)+3A-1\right)
   \left(G^3\left(1+\frac{3}{H}\right)+3A+1\right)
\end{align}
$$
Then we have either
$$
1-\frac{3}{H} &lt; 0 \\
\implies H &lt; 3 \implies
1&lt; \frac{1}{x}+\frac{1}{y}+\frac{1}{z}
$$
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
$$
\begin{align}
8^3 &amp;= \left(G^3\left(1-\frac{3}{H}\right)+3A-1\right)
   \left(G^3\left(1+\frac{3}{H}\right)+3A+1\right) \\
&amp;\ge (H^3-3H^2+3H-1)(H^3+3H^2+3H+1) \\
&amp;= (H-1)^3(H+1)^3 \\
&amp;= (H^2-1)^3 \\
\implies H &amp;\le 3 \\
\implies 1 &amp;\le \frac{1}{x}+\frac{1}{y}+\frac{1}{z}
\end{align}
$$
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 $1$.