Take any prime dividing $a,b$ or $c$. Let $p^k \| a, p^m \|b, p^n \|c$. Then the conditions imply that $k \le 3m, m \le 3n, n \le 3k$. W.l.o.g. let $k \ge m,n$. Then certainly $p^{13k} \mid (a+b+c)^{13}$ whereas $p^{k+m+n} \| abc$ so it suffices to prove that $k+m+n \le 13k$ which is obvious since $k \le k, m \le 9k, n \le 3k$. Done.

Just apply multinomial theorem for $(a+b+c)^{13}$ and use the given data.This problem becomes trivial after application of multinomial theorem.

andersarnold123 wrote: Just apply multinomial theorem for $(a+b+c)^{13}$ and use the given data.This problem becomes trivial after application of multinomial theorem. You are welcome to show us a complete solution shorter than the one presented in #2.

There is definitely a solution using multinomial expansion. However, it is much longer as compared to the solution in #2.

a genral way of solving: lets notice that all the elements in $(a+b+c)^{13}$ are in the form of $a^ib^jc^k$, s.t $i+j+k=13$. we will show that every one of them is divisible by abc: lets divide to cases: case 1: $i,j,k\ge 1$, which is simple to show. case 2: one of $i,j,k$ is equal to zero, without loss generality lets assume $k=0$. so it can be written as $a^i*b^j$ s.t $i+j=13$. sub cases are that or $i\ge 10$ or $j\ge 4$, and in both of them we can $a*a^{i-1}*b*b^{j-1}$, and or $a^{i-1}$ is at least $a^9$ and divisible by $b^3$ and therefore also by c, or $b^{j-1}$ is at least $b^3$ and divisible by c. case 3: two of $i,j,k$ is equal to zero, without loss generality lets assume $j,k=0$.so it can be written as $a^i$, and actually as $a^{13}$, which can be written $a*a^3*a^9$, which $b|a^3$ and $c|b^3$ and $b^3|a^9$ so also divided by c. after all, all the elements by themselves is divisible by $abc$ and therefor the expression itself

note that this question is almost same as RMO 2002, problem 3.