Prove $\text{lcm}(a,b,c) = \frac{a \cdot b \cdot c \cdot \gcd(a,b,c)}{\gcd(a,b)\gcd(b,c)\gcd(a,c)}$

Question

With $\{a,b,c\}\subset \mathbb{Z}$ given that $\text{lcm}(a,b) = \frac{a \cdot b}{\gcd(a,b)}$. Prove that: $$\text{lcm}(a,b,c) = \frac{a \cdot b \cdot c \cdot \gcd(a,b,c)}{\gcd(a,b)\gcd(b,c)\gcd(a,c)}$$

Answer 1

Let $a=p_1^{\alpha_1}p_2^{\alpha_2}...p_n^{\alpha_n}$, $b=p_1^{\beta_1}p_2^{\beta_2}...p_n^{\beta_n}$ and $c=p_1^{\gamma_1}p_2^{\gamma_2}...p_n^{\gamma_n}$, where $\alpha_i\geq0$, $\beta_i\geq0$, $\gamma_i\geq0$ be integers and $p_i$ be different prime numbers.

Hence, we need to prove that $$p_1^{\max\{\alpha_1,\beta_1,\gamma_1\}}p_2^{\max\{\alpha_2,\beta_2,\gamma_2\}}...p_n^{\max\{\alpha_n,\beta_n,\gamma_n\}}=$$ $$=p_1^{\alpha_1+\beta_1+\gamma_1+\min\{\alpha_1,\beta_1,\gamma_1\}-\min\{\alpha_1,\beta_1\}-\min\{\alpha_1,\gamma_1\}-\min\{\beta_1,\gamma_1\}}\cdot...$$ $$\cdot p_n^{\alpha_n+\beta_n+\gamma_n+\min\{\alpha_n,\beta_n,\gamma_n\}-\min\{\alpha_n,\beta_n\}-\min\{\alpha_n,\gamma_n\}-\min\{\beta_n,\gamma_n\}}.$$

Thus, it's enough to prove that $$\max\{x,y,z\}+\min\{x,y\}+\min\{x,z\}+\min\{y,z\}=x+y+z+\min\{x,y,z\},$$ which is obvious because we can assume $x\geq y\geq z$.

Done!