Problem 1735

Let $x = \frac{2a}{a-b}$, $y = \frac{2b}{b-c}$, $z = \frac{2c}{c-a}$. Then
$$(1 - \frac{2}{x}) (1 - \frac{2}{y}) (1 - \frac{2}{z}) = 1,$$
which is equivalent to
$$xy + yz + zx - 2(x + y + z) + 4 = 0.$$
If we denote $x + y + z = s$, then the inequality of the problem becomes
$$\begin{align*} & x^2 + y^2 + z^2 - 4(x + y + z) + k \ge 0 \\ \iff & s^2 - 2(xy + yz + zx) - 4s + k \ge 0 \\ \iff & s^2 - 2(2s - 4) - 4s + k = (s - 4)^2 + k - 8 \ge 0. \end{align*}$$
The right hand side of the latest equality is sum of a complete square and $k - 8$ thus we must have $8 \le k$ so that the inequality always holds. Equality holds when $x + y + z = 4$. One possible triple which satisfies $x + y + z = 4$ and $xy + yz + zx - 2(x + y + z) + 4 = 0$ is $(x, y, z) = (\frac{8}{3}, \frac{2}{3}, \frac{2}{3})$, which is attained for $(a, b, c) = (4, 1, -2)$ so we can conclude that the desired number is $k = 8$.