$a\mid b^2, b\mid c^2, c\mid a^2\Longrightarrow abc\mid (a+b+c)^k$ Is it true that if $a,b,c$ are positive integers such that $a\mid b^2, b\mid c^2, c\mid a^2$, then $abc$ always divides $(a+b+c)^6$?
If not (I couldn’t find a counterexample), then $abc\mid (a+b+c)^7$ is always true? Or in general, find the minimum value of $k$, such that $abc\mid (a+b+c)^k$.
I proved that $a,b,c\mid (a+b+c)^6$, but unfortunetly that is not enough, since $a,b,c$ are not necessarily relative primes.

The minimum qualifying value of $k$ is $7$.

To show that $k=6$ is not sufficient, let $p$ be a prime.

Then letting $(a,b,c)=(p,p^4,p^2)$, we get 
\begin{align*}
abc&=p^7\\[4pt]
(a+b+c)^6&=p^6(1+p^3+p)^6\\[4pt]
\end{align*}
from which it follows that $abc$ does not divide $(a+b+c)^6$.

Next we show that $k=7$ always works . . .

Let $a,b,c$ be positive integers such that
$$a|b^2,\;\;\;b|c^2,\;\;\;c|a^2$$
Claim:$\;abc$ divides $(a+b+c)^7$.

If any of $a,b,c$ equals $1$, then they must all be equal to $1$, in which case the truth of  the claim is immediate.

Thus, assume $a,b,c > 1$.

From the given divisibility conditions on $a,b,c$, it follows that $a,b,c$ have the same set of distinct prime factors.

Let $p$ be an arbitrary prime factor of $abc$.

It follows that $p$ is a common prime factor of $a,b,c$.

Letting $i,j,k$ be positive integers such that
$$p^i||a,\;\;\;p^j||b,\;\;\;p^k||c$$
it follows that $p^{i+j+k}{\,||\,}abc$, and $p^{7\min(i,j,k)}{\;\mid\,}(a+b+c)^7$.

Using the given divisibility conditions,
\begin{align*}
a|b^2\implies i\le 2j\\[4pt]
b|c^2\implies j\le 2k\\[4pt]
c|a^2\implies k\le 2i\\[4pt]
\end{align*}
so we get
\begin{align*}
{\small{\bullet}}\;\,&i+j+k\le i+(2k)+k=i+3k\le i+6i=7i\\[4pt]
{\small{\bullet}}\;\,&i+j+k\le i+j+(2i)=3i+j \le 6j+j=7j\\[4pt]
{\small{\bullet}}\;\,&i+j+k\le (2j)+j+k=3j+k\le 6k+k=7k\\[4pt]
\end{align*}
hence $i+j+k\le 7\min(i,j,k)$.

Thus the highest power of $p$ which divides $abc$ is less than or equal to the highest power of $p$ which divides $(a+b+c)^7$.

Since $p$ was an arbitrary prime factor of $abc$, it follows that $abc$ divides $(a+b+c)^7$, as claimed.
