Does this sort of triangle look familiar to you?

For any triangle

Radius of incircle $\left( r \right) = \dfrac{\varDelta}{s}$

where $s$ is the semi-perimeter of $\Delta ABC$ , i.e., $s = \dfrac{a+b+c}{2}$ .

So

$\begin{aligned} & \dfrac 6s = 1 \\ \implies & s = 6 \end{aligned}$

Using heron's formula

$\begin{aligned} & \varDelta = \sqrt{s(s-a)(s-b)(s-c)} \\ \implies & 6 = \sqrt{6(6-a)(6-b)(6-c)} \\ \implies & 36 = 6(6-a)(6-b)(6-c) \\ \implies & 6 = 216 - 36(a+b+c) + 6(ab+bc+ca) - abc \\ \implies & 6 = 216 - 36(12) + 6(ab+bc+ca) - 60 \qquad \qquad \qquad \qquad \small \color{#3D99F6}{\because a+b+c = 2s = 12} \\ \implies & ab+bc+ca = 47 \end{aligned}$

Now

$\dfrac 1a + \dfrac 1b + \dfrac 1c = \dfrac{ab+bc+ca}{abc} = \dfrac{47}{60} \approx \boxed{0.783}$

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Extra bit:

An interesting thing to note here is that the only unordered triplet for positive integral values of $a,b$ and $c$ is $(a,b,c)=(3,4,5)$ which is a very popular Pythagorean triplet! (Thus, the title is justified.)