If $a+b+c$ divides the product $abc$, then is $(a,b,c)$ a Pythagorean Triple?

Firstly, I will define what Pythagorean Triples are for those who do not know.

Definition:

A Pythagorean Triple is a group of three integers $a$, $b$ and $c$ such that $a^2+b^2=c^2$, since the Pythagorean Theorem asserts that for any $90^\circ$ (right-angle) triangle $ABC$ with sides $a$, $b$ and $c$, one will always have the equation, $a^2+b^2=c^2$.

I was looking at Pythagorean Triples and noticed another property apart from how $a^2+b^2=c^2$. Here are the first $30$ Pythagorean Triples $(a,b,c)$ ordered from smallest to greatest value, i.e. $$(a,b,c)\qquad\text{ s.t. }\qquad a<b<c.\tag*{$\big(\text{s.t. = such that}\big)$}$$

I noticed that $a^2=(c+b)(c-b)$, but that is trivial since $$\begin{align}a^2&=(c+b)(c-b)\tag{given} \\ &=c^2-b^2 \\ \Leftrightarrow\,\,\,\, a^2+b^2&=c^2.\end{align}$$

However, I also noticed that by having "$u\mid v$" be read as "$u$ divides $v$", it appears that $$a+b+c\mid abc.$$ For example, $(a,b,c)=(3,4,5)$ is a classic Pythagorean Triple; $3^2+4^2=5^2$.

Also, $$\begin{align}3+4+5&=12 \\ \& \quad3\times 4\times 5 &= 60. \\ \\ 12 &\,\mid 60 \\ \Leftrightarrow \,\,\,\,3+4+5&\,\mid 3\times 4\times 5.\end{align}$$ This, I cannot prove to be true $-$ but I tested with all the $30$ Pythagorean Triples above, and I have come across no counter-example. Is there a proof? I do not know where to begin myself.

Conjecture:

Given three positive integers $a$, $b$ and $c$, if $a < b<c$ and $a^2+b^2=c^2$, then $$a+b+c\mid abc.$$

Thank you in advance.

Edit:

My conjecture was originally the other way round; i.e. if $a+b+c\mid abc$ then $a^2+b^2=c^2$. But $6$ is a counter-example, namely because it is a Perfect Number.

You actually want it the other way around: if $a^2+b^2=c^2$ then $a+b+c|abc$. That you can prove very quickly from the general form of primitive Pythagorean triples $(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$.

Alternatively: $$\frac{abc}{a+b+c}=\frac{abc(a+b-c)}{(a+b+c)(a+b-c)}=\frac{abc(a+b-c)}{2ab}=\frac{c(a+b-c)}{2},$$ which is a positive integer for both cases: $a,b,c$ are all even; $a,c$ are odd and $b$ is even.