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.
 A: 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
$$
\begin{align}
8^3 &= \left(G^3\left(1-\frac{3}{H}\right)+3A-1\right)
   \left(G^3\left(1+\frac{3}{H}\right)+3A+1\right) \\
&\ge (H^3-3H^2+3H-1)(H^3+3H^2+3H+1) \\
&= (H-1)^3(H+1)^3 \\
&= (H^2-1)^3 \\
\implies H &\le 3 \\
\implies 1 &\le \frac{1}{x}+\frac{1}{y}+\frac{1}{z}
\end{align}
$$
A: Let $x^2-1=\frac{8bc}{a^2}$, $y^2-1=\frac{8ac}{b^2}$ and $z^2-1=\frac{8ab}{c^2}$, where $a$, $b$ and $c$ are positives.
Hence, by Holder we obtain:
$$\sum_{cyc}\frac{1}{x}=\sum_{cyc}\frac{1}{\sqrt{1+\frac{8bc}{a^2}}}=\sum_{cyc}\frac{a}{\sqrt{a^2+8bc}}=$$
$$=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{a}{\sqrt{a^2+8bc}}\right)^2\sum\limits_{cyc}a(a^2+8bc)}{\sum\limits_{cyc}a(a^2+8bc)}}\geq\sqrt{\frac{(a+b+c)^3}{\sum\limits_{cyc}a(a^2+8bc)}}=$$
$$=\sqrt{\frac{\sum\limits_{cyc}(a^3+3a^2b+3a^2c+2abc)}{\sum\limits_{cyc}(a^3+8abc)}}=\sqrt{\frac{\sum\limits_{cyc}(a^3+8abc)+3\sum\limits_{cyc}c(a-b)^2}{\sum\limits_{cyc}(a^3+8abc)}}\geq1.$$
Done!
A: 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 $1$.
