Prove that: $\gcd[a,b,c]=\frac{abc.\operatorname{lcm}(a,b,c)}{\operatorname{lcm}(a,b)\operatorname{lcm}(a,c)\operatorname{lcm}(b,c)}$

Question

I may be wrong, but I was thinking of:

$\operatorname{lcm}(a, b)$ as $\min(a, b)$ and $\gcd (a, b)$ as $\max(a, b)$ and $a <b <c$

I know I'm wrong, but I think I can do it

Answer 1

No. lcm means least common multiple and gcd means greatest common divisor.

Let lcm $(a,b,c)=R$

Let lcm $(a,b)=RS$

Let lcm $(a,c)=RT$

Let lcm $(b,c)=RU$

Then there are further integers $A,B,C$ such that $a=RSTA,b=RSUB,c=RTUC$. Furthermore, gcd $(a,b,c)=RSTUABC$

Plugging these into the formula is now straightforward.

Answer 2

\begin{align*} \gcd(a,b,c)&=\gcd(a,\gcd(b,c)) \\ &= \frac{a\cdot\gcd(b,c)}{\mathrm{lcm}(a,\gcd(b,c))}\\ &= \frac{a\cdot\gcd(b,c)}{\gcd(\mathrm{lcm}(a,b)\mathrm{lcm}(a,c))}\\ &= \frac{a\cdot bc/\mathrm{lcm}(b,c)}{\mathrm{lcm}(a,b)\mathrm{lcm}(a,c)/\mathrm{lcm}(\mathrm{lcm}(a,b),\mathrm{lcm}(a,c))}\\ &= \frac{abc\cdot\mathrm{lcm}(\mathrm{lcm}(a,b),\mathrm{lcm}(a,c))}{\mathrm{lcm}(b,c)\mathrm{lcm}(a,b)\mathrm{lcm}(a,c)}\\ &= \frac{abc\cdot\mathrm{lcm}(a,b,c)}{\mathrm{lcm}(b,c)\mathrm{lcm}(a,b)\mathrm{lcm}(a,c)}. \end{align*}

Answer 3

Not the most elegant solution but pretty straight forward let $a=p_1^{k_1}\cdot ...\cdot p_s^{k_s}$, $b=p_1^{l_1}\cdot ...\cdot p_s^{l_s}$ and $c=p_1^{m_1}\cdot ...\cdot p_s^{m_s}$, where $k,l,m,s$ are non-negative integers. Now lets compare $p_i$ on both sides. Let $k_i \leq l_i \leq m_i$ then on the left side we get $p_i^{k_i}$ (because $\min(k_i,\min(l_i,m_i))=\min(k_i,l_i)=k_i$). On the right side we get $p_i^{k_i+l_i+m_i+\max(k_i,\max(l_i,m_i))-\max(k_i,l_i)-\max(l_i,m_i)-\max(k_i,m_i)}=p_i^{k_i+l_i+m_i+m_i-l_i-m_i-m_i}=p_i^{k_i}$. Thus the equality holds.