<s>[INST] 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.
 [/INST] $x^2=a^2-1,y^2=b^2-1,z^2=c^2-1 \implies (x^2+1)(y^2+1)(z^2+1)\ge 64 \to x+y+z \ge 3\sqrt{3}$
let $3u=x+y+z,3v^2=xy+yz+xz,w^3=xyz \implies u \ge v \ge w,(x^2+1)(y^2+1)(z^2+1)\ge 64 \iff w^6-6uw^3+9u^2-6v^2+9v^4-63=f(w^3) \ge 0  \implies \Delta\le 0 \implies 36u^2-4(9u^2-6v^2+9v^4-63) \le 0 \iff 3v^4-2v^2-63 \ge 0 \iff (v^2-3)(3v^2+7) \ge 0 \implies v^2 \ge 3 \implies u^2 \ge v^2 \ge 3 \implies x+y+z \ge 3\sqrt{3}$
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