Problem 2212

Answer $(1, 1, 1)$, $(1, 2, 2)$, $(2, 3, 6)$.
Let $p = a + b + c$, then $a + b = p - c$, $b + c = p - a$, $c + a = p - b$ and
$$\frac{(p-c)^4}{c} + \frac{(p-a)^4}{a} + \frac{(p-b)^4}{b}$$
is a non-negative integer. By expanding brackets we obtain that the number $p^4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ is integer, too. But the numbers $a$, $b$, $c$ are not divisible by $p$, therefore the number $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is (non negative) integer. That is possible for the triples $(1, 1, 1)$, $(1, 2, 2)$, $(2, 3, 6)$ only.