Solution:
The crux of this problem is to apply the trigonometric substitutions $a = \cot \alpha$, $b = \cot \beta$, and $c = \cot \gamma$, with $0 < \alpha, \beta, \gamma < \pi / 2$. Then, the given equations translate to
$$\frac{2}{\sin^{2}\alpha} = \frac{3}{\sin^{2}\beta} = \frac{4}{\sin^{2}\gamma} \quad \mathrm{and} \quad \cot \alpha \cot \beta + \cot \beta \cot \gamma + \cot \gamma \cot \alpha = 1.$$
From the second equation, we get
$$\cot \gamma = \frac{1 - \cot\alpha\cot\beta}{\cot\alpha + \cot\beta} = -\cot (\alpha + \beta).$$
Since $\alpha$, $\beta$, and $\gamma$ all between $0$ and $\pi / 2$, we discover that
$$\alpha + \beta + \gamma = \pi.$$
Let $\triangle ABC$ be the (acute) triangle with side lengths $BC = \sqrt{2}$, $CA = \sqrt{3}$, and $AB = \sqrt{4}$. By Law of Sines, setting $\alpha = \angle A$, $\beta = \angle B$, and $\gamma = \angle C$ will satisfy both equations. Thus, Law of Cosines gives
$$\cos \alpha = \frac{3 + 4 - 2}{2\cdot\sqrt{3}\cdot\sqrt{4}} = \frac{5}{\sqrt{48}} \Rightarrow a = \cot \alpha = \frac{5}{\sqrt{23}}$$
Similar calculations give $b = \frac{3}{\sqrt{23}}$ and $c = \frac{1}{\sqrt{23}}$, so the answer is $a + b + c = \left[\frac{9}{\sqrt{23}}\right]$.

Solution:
Let $2(a^{2} + 1) = 3(b^{2} + 1) = 4(c^{2} + 1) = x$. Then, since $ab + bc + ca = 1$, we have the following system of equations:
$$(a + b)(c + a) = a^{2} + ab + bc + ca = a^{2} + 1 = x / 2$$
$$(b + c)(a + b) = b^{2} + ab + bc + ca = b^{2} + 1 = x / 3$$
$$(c + a)(b + c) = c^{2} + ab + bc + ca = c^{2} + 1 = x / 4.$$
Taking advantage of symmetry, we discover that
$$a + b = \sqrt{\frac{2x}{3}},\quad b + c = \sqrt{\frac{x}{6}},\quad \mathrm{and} \quad c + a = \sqrt{\frac{3x}{8}}.$$
To solve for $x$, notice that
$$2 = 2(ab + bc + ca)$$
$$\quad = (a + b)^{2} + (b + c)^{2} + (c + a)^{2} - 2(a^{2} + b^{2} + c^{2})$$
$$\quad = \frac{2x}{3} + \frac{x}{6} + \frac{3x}{8} - 2\left(\frac{x}{2} - 1 + \frac{x}{3} - 1 + \frac{x}{4} - 1\right)$$
$$\quad = -\frac{23x}{24} + 6,$$
so $x = \frac{96}{23}$. Therefore,
$$a + b + c = \frac{1}{2}\left(\sqrt{\frac{2x}{3}} + \sqrt{\frac{x}{6}} + \sqrt{\frac{3x}{8}}\right) = \frac{1}{2}\left(\frac{8 + 4 + 6}{\sqrt{23}}\right) = \left[\frac{9}{\sqrt{23}}\right].$$