Relationship between GCD and LCM

Question

Could someone help me prove that $$\frac{1}{{\gcd \left( {\frac{1}{{{a_1}}}, \ldots ,\frac{1}{{{a_n}}}} \right)}} = \operatorname{lcm} \left( {{a_1}, \ldots ,{a_n}} \right) $$ without using that $\gcd \left( {a,b} \right)\operatorname{lcm} \left( {a,b} \right) = ab$.

Answer 1

As in a closely related question on gcd & lcm of rationals

$\begin{align} {\rm lcm}(a_1,\ldots,a_n)\mid x \iff\ \ &a_1,\ldots a_n\mid x\\ \iff\ \ &\ 1\mid x/a_1,\ldots, x/a_n\\ \iff\ \ &\ 1\mid \gcd(x/a_1,\ldots, x/a_n)\\ \iff\ \ &\ 1\mid x\,\gcd(1/a_1,\ldots, 1/a_n) =: x\,g\\ \iff\ \ &\!\!\!\!1/g\mid x \end{align}$