Prove that LCM$(\frac{n}{a}, \frac{n}{b}) = n$ if $(a,b)=1$.
Let us assume the factorization of $n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$ and $a=q_1^{\beta_1} q_2^{\beta_2} \cdots q_k^{\beta_k}$ and $b = r_1^{\gamma_1} r_2^{\gamma_2} \cdots p_k^{\gamma_k}$.
Please note that both $a$ and $b$ divide $n$.
Case 1: If both $a$ and $b$ divide $n$ then I am getting that $\frac{n}{a}$ will divide $n$ and $\frac{n}{b}$ will also divide $n$ but why it will be the minimum multiple?
On
Suppose $n/a$ divides $x$ and $n/b$ divides $m$, say $m=xn/a=yn/b$, for some integers $x,y$.
Then we have $xb=ya$. Since $a$ is co-prime to $b$, this implies $a$ divides $x$; similarly $b$ divides $y$. Consequently $m=xn/a=n\cdot x/a$ is a multiple of $n$.
Thus $n$ is the least common multiple of $n/a$ and $n/b$.
Hope this helps.
I find it easier to work with gcd. The following 2 lemmas are useful and not hard to prove.
Lemma 1. If $ (a,b) = 1 $, then $ (ak,bk) = k $ for any integer $ k $.
Lemma 2. $ \gcd(a,b) \text{ lcm}(a,b) = ab $.
Given these two results, we can now prove as follows: let $ k = \frac{n}{ab} $ in Lemma 1. Note that $ ab $ divides $ n $ since $ a $ and $ b $ are coprime. Then $ \gcd(\frac{n}{a}, \frac{n}{b} ) = ( \frac{bn}{ab}, \frac{an}{ab}) = \frac{n}{ab} $. Now use lemma 2: $$ \frac{n^2}{ab} = \frac{n}{a} \frac{n}{b} = \gcd(\frac{n}{a}, \frac{n}{b}) \text{lcm}(\frac{n}{a}, \frac{n}{b}), $$ and since $ \gcd = \frac{n}{ab} $ we get $ \text{lcm} = n $.