Number Theory Least Common Multiple Question

I don't know how I could find, WITH proof, the smallest possible least common multiple of three positive integers which sum to $2005$.

If someone can provide a proof I would be very happy.

Solution 1:

$2005=4\cdot401$ with $401$ prime.

Then

$$2005=2\cdot401+2\cdot401+401 \,.$$

Thus we found three numbers with lcm 802. We claim that this is the smallest possible value.

Suppose by contradiction that we can find

$a+b+c =2005$ and $l=\operatorname{lcm}(a,b,c) < 802$. We can assume that $a \leq b \leq c$.

Since $b, c \leq l$ it follows that

$$ 2005=a+b+c \leq a+2l < a+2\cdot802$$

Thus $a > 401> \frac{l}{2}$. Since $a> \frac{l}{2}$ and $a\mid l$, it follows that $a=l$.

Then $$l =a \leq b \leq c \leq l \Rightarrow a=b=c=l \Rightarrow 3l=2005$$

Contradiction.