When one of the numbers is zero, say $c=0$, we have $a^{3k-1}b \leq (3k-1)^{3k-1}$, with equality for $b=1$, $a=3k-1$, by AM-GM, since then $3k = a + b = (3k-1)\frac {a}{3k-1} + b \geq 3k\sqrt[3k]{\frac {a^{3k-1}b} {(3k-1)^{3k-1}}}$. They say that by choosing the middle number from $a,b,c$, say $b$, one can prove that $b(a+c)^{3k-1} \geq a^{3k-1}b+b^{3k-1}c+c^{3k-1}a+k^{2}a^{k}b^{k}c^{k}$, and now we will be under the condition $a'= a+c$, $b'= b$, $c' = 0$ solved above. EDIT. Oh, this wasn't meant to be a proof I only computed the maximum value on the border of the domain of definition, and just quoted by hearsay the way to reduce it to the border.

This inequality is generalization of a well known one: $a^2b+b^2c+c^2a+abc \leq \frac{4}{27}(a+b+c)^3$ and it can be proved in the same way, so in my opinion, it's not the best choice for such an important tst.. This is my proof (basically the same as mavropnevma proof with few things explained more): WLOG we can suppose that $a=max \{a,b,c \}$. We can see that for $a\ge b\ge c$ we have: $a^n b +b^n c+ c^n a \geq ab^n+bc^n+ca^n$, so we can suppose that $a\ge b\ge c$. Denote now $f(a,b,c)=a^{3k-1}b+b^{3k-1}c+c^{3k-1}a + k^2 a^kb^kc^k$ We want to prove that $f(a,b,c) \leq f(a+c,b,0)$, what is equal to: $(a+c)^{3k-1}b \geq a^{3k-1}b+b^{3k-1}c+c^{3k-1}a + k^2 a^kb^kc^k \quad (1)$ We can see that when we expand LHS, we will have: $a^{3k-1}b+a^{3k-2}bc+(3k-2)a^{3k-2}bc+ \dbinom{3k-1}{2} a^{3k-3}bc^2+ ...+(3k-1)c^{3k-2}ab$ Now we can see that for $k=1$, $(1)$ follows from first three and the last summands, and for $k\ge 2$, $(1)$ follows from the first four summands because $\dbinom{3k-1}{2} \geq k^2$. Hence, maximum is reached when $c=0$, and it is equal to $f(a,b,c) \leq (3k-1)^{3k-1}$, what mavropnevma showed. Equality holds for $k=1$ when $(a,b,c)=(1,1,1)$ or $(2,1,0)$ and for $k\ge 2$ when $(a,b,c)=(3k-1,1,0)$ with cyclic permutations. $\blacksquare$

This inequality was proposed by me for the IMO 2011 in Netherlands, and it was not selected because it's a generalization of the well known inequality written above. Greece proposed this again for Balkan Olympiad 2012, so maybe they took it from there...

Indeed, they took the problem from the BMO shortlist.