Given $a,b,c\ge1;abc\ge8$. Proving that $\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$ 
Given $a,b,c\ge1;abc\ge8$. Proving that $$\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$$

I have tried by using Jensen's inequality:
We consider the inequality: $\displaystyle\sqrt{x^2-1}\ge\sqrt3+\frac4{\sqrt3}\ln\frac x2\tag{i}$
When $x>\frac85$, I can prove that $(\text i)$ true. But it is clear that $(\text i)$ is not true for all $x<\frac85$, and I can't prove that $\displaystyle\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$ in this case.
 A: $x^2=a^2-1,y^2=b^2-1,z^2=c^2-1 \implies (x^2+1)(y^2+1)(z^2+1)\ge 64 \to x+y+z \ge 3\sqrt{3}$
let $3u=x+y+z,3v^2=xy+yz+xz,w^3=xyz \implies u \ge v \ge w,(x^2+1)(y^2+1)(z^2+1)\ge 64 \iff w^6-6uw^3+9u^2-6v^2+9v^4-63=f(w^3) \ge 0  \implies \Delta\le 0 \implies 36u^2-4(9u^2-6v^2+9v^4-63) \le 0 \iff 3v^4-2v^2-63 \ge 0 \iff (v^2-3)(3v^2+7) \ge 0 \implies v^2 \ge 3 \implies u^2 \ge v^2 \ge 3 \implies x+y+z \ge 3\sqrt{3}$
A: Set $a=e^x,b=e^y,c=e^z$. Then we have to prove:
$$ f(x,y,z)=\sum_{cyc}\sqrt{e^{2x}-1}\geq 3\sqrt{3}\tag{1} $$
with the constraints $x,y,z\geq 0$ and $x+y+z= 3\log 2$. Obviously $x=y=z=\log 2$ is a stationary point (a local minimum) for $f$ over the given domain and $f(\log 2,\log 2,\log 2)=3\sqrt{3}$. 
The solutions of
$$ \frac{e^{2x}}{\sqrt{e^{2x}-1}}=\lambda \tag{2}$$
for $\lambda\geq 2$, are given by $e^x=\frac{1}{\sqrt{2}}\sqrt{\lambda^2\pm \lambda\sqrt{\lambda^2-4}}$, leading to $\sqrt{e^{2x}-1}=\frac{\lambda\pm\sqrt{\lambda^2-4}}{2}$. By studying all the possible chances  $(\pm,\pm,\pm)$ (they are not many, up to symmetry) is is not difficult to see that $(x,y,z)=(\log 2)\cdot(1,1,1)$ is a global minimum for the interior of the triangle. After that, we just have to study the boundary of the triangle (one side is enough, always by symmetry) but that is the same problem with just two variables, way easier. By putting things together, we have that the previous point is a global minimum as wanted.
