Seeing the condition $a+b+c=abc$ reminds us of the trigonometric substitution $a=\tan A$ , $b=\tan B$ and $c=\tan C$ where $A+B+C=\pi$ .

Subbing this back in the expression, we get $\sqrt{(\tan^2 A+1)(\tan^2B+1)(\tan^2C+1)}$ Using the identity $\tan^2\theta + 1 = \sec^2\theta$ we get it equal to $\sqrt{(\sec^2 A)(\sec^2B)(\sec^2C)}=\sec A\sec B\sec C=\dfrac{1}{\cos A\cos B\cos C}$

However, we know that $\cos A\cos B\cos C\le \dfrac{1}{8}$ as it is a famous inequality, so $\dfrac{1}{\cos A\cos B\cos C}\ge 8$ and our answer is $\lfloor 8\cdot 100\rfloor =\boxed{800}$

I absolutely admire Daniel's approach. But as Samuel wanted an algebraic one, here it is... :)

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First, let $m = a+b+c = abc, n = ab+bc+ca,$ and $f(x) = (x+a)(x+b)(x+c)$ .

If we expand the formula for $f$ , we get $f(x) = x^3 + mx^2 + nx + m$ . Therefore, we have: $\left\{ \begin{array}{ll} f(i) = i^3+ni + m(i^2+1) = -i + ni = (n-1)i\\ f(-i) = (-i)^3 - ni + m(i^2+1) = i - ni = -(n-1)i\end{array} \right.$ (where $i = \sqrt{-1}$ .) Therefore, $\sqrt{(a^2+1)(b^2+1)(c^2+1)} = \sqrt{(a+i)(b+i)(c+i)(a-i)(b-i)(c-i)}$ $= \sqrt{f(i)f(-i)} = \sqrt{-i^2(n-1)^2} = \sqrt{(n-1)^2} = |n-1|$ Now let's use the $AM-GM$ inequality to find a lower bound on $n$ : $n = ab+bc+ca \ge 3\left(\sqrt[3]{a^2 b^2 c^2}\right) = 3m^{2/3}$ Since a few people have already proved that $abc \ge 3\sqrt{3}$ , I won't bother proving it again. Substituting this gives us $n \ge 9$ , and therefore $|n-1| \ge \boxed{8}$ .

Actually, i used another approach but not sure maybe it's mathematicaly acceptable.

$a+b+c=abc$ ,

therefore, $\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=1$ ,

using $AM-GM$ ,

$\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=1 \geq$ $3 3\sqrt{\frac{1}{a^{2}b^{2}c^{2}}}$ ,

equality holds when $a=b=c$

therefore, $1 \geq 3 3\sqrt{\frac{1}{a^{6}}}$

$1 \geq 3a^{-2}$ ,

$a^{2} \geq 3$ and

$a \geq \sqrt{3}$ .

$a,b,c \geq \sqrt{3}$

Subtituting into the inequality,

we get $m = \sqrt{64} =8$ .

Therefore, floor function of $100m= 800$ .

I don't know maybe this approach is correct.

$(a^2+1)(b^2+1)(c^2+1)$

$=1+(abc)^2(1+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})$

$\ge 1+(abc)^2(1+3(abc)^{-2/3})$ (AM-GM)

$=1+(abc)^2+3(abc)^{4/3}$

Which is increasing with abc. Thus if equality holds when abc is minimized, then this is minimized.

$((a+b+c)/3)^3 \ge abc \\((abc)/3)^3 \ge abc \\abc \ge 3\sqrt{3}$

With equality when $a=b=c=\sqrt{3}$ . As these values minimize abc and also hold the equality, they give the minimum value of m, i.e. $\sqrt{(\sqrt{3}^2+1)^3}=8$

Using AM-GM we get $a+b+c\geq3\sqrt[3]{abc}$ $\Leftrightarrow abc\geq3\sqrt[3]{abc}$ $\Leftrightarrow \sqrt[3]{(abc)^2}\geq3$ Using a consequense of Holder inequality we get $(a^2+1)(b^2+1)(c^2+1)\geq[1+\sqrt[3]{(abc)^2}]^3\geq4^3$ $\Leftrightarrow\sqrt{(a^2+1)(b^2+1)(c^2+1)}\geq\sqrt{4^3}=8$ $\Rightarrow m=8\Rightarrow \lfloor 100m\rfloor =800$ The equality holds when $a=b=c=\sqrt{3}$