Is $\mbox{lcm}(a,b,c)=\mbox{lcm}(\mbox{lcm}(a,b),c)$? $\newcommand{\lcm}{\operatorname{lcm}}$Is $\lcm(a,b,c)=\lcm(\lcm(a,b),c)$?
I managed to show thus far, that $a,b,c\mid\lcm(\lcm(a,b),c)$, yet I'm unable to prove, that $\lcm(\lcm(a,b),c)$ is the lowest such number... 


Lemma:
Let $l:=\operatorname{lcm}(a,b,\dots,z)$.
Let $m$ be a common multiple of $a,b,\dots z$.
Then, $m$ is a multiple of $l$.
Proof:
We can write $m=ql+r$, $0\leq r<l$.
Then, $r=m-ql$.
Both $m$ and $l$ are a multiple of $a$.
So, $r$ is a multiple of $a$.
Both $m$ and $l$ are a multiple of $b$.
So, $r$ is a multiple of $b$.
$\dots$
Both $m$ and $l$ are a multiple of $z$.
So, $r$ is a multiple of $z$.
So, $r$ is a common multiple of $a,b\dots, z$.
If $0<r<l$, then $r$ is a common multiple of $a,b\dots, z$ which is less than the least common multiple of $a,b\dots, z$.
This is a contradiction.
So, $r$ must be $0$.
So, $m=ql+r=ql$.
So, $m$ is a multiple of $l$.


$\operatorname{lcm}(a,b,c)=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$.
Proof:
Let $l:=\operatorname{lcm}(a,b,c)$ and $l^{'}:=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$.
$l$ is a common multiple of $a$, $b$ and $c$.
So, $l$ is a common multiple of $a$ and $b$.
So, by the above lemma, $l$ is a multiple of $\operatorname{lcm}(a,b)$.
$l$ is a multiple of $c$.
So, $l$ is a common multiple of $\operatorname{lcm}(a,b)$ and $c$.
So, by the above lemma, $l$ is a multiple of $l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$.
$l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a multiple of $\operatorname{lcm}(a,b)$.
$\operatorname{lcm}(a,b)$ is a multiple of $a$.
So, $l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a multiple of $a$.
$l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a multiple of $\operatorname{lcm}(a,b)$.
$\operatorname{lcm}(a,b)$ is a multiple of $b$.
So, $l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a multiple of $b$.
$l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a multiple of $c$.
So, $l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a common multiple of $a$,$b$ and $c$.
So, by the above lemma, $l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a multiple of $l=\operatorname{lcm}(a,b,c)$.
$l$ is a multiple of $l^{'}$ and $l^{'}$ is a multiple of $l$.
So, $l=l^{'}$.

