The symbols $(a,b,\ldots,g)$ and $[a,b,\ldots, g]$ denote the greatest common divisor and least common multiple, respectively, of the positive integers $a,b,\ldots, g$. For example, $(3,6,18)=3$ and $[6,15]=30$. Prove that
$\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}$.
USAMO 1972/1
"Questions we can't answer are far better than answers we can't question"
Re: USAMO 1972/1
i think it is a very familiar problem
use PPF of a,b,c and work with the power of primes .
use PPF of a,b,c and work with the power of primes .
Re: USAMO 1972/1
$\text{My Proof:}$
Let $p = (a, b, c)$, $pq = (a, b)$, $pr = (b, c)$, and $ps = (c, a)$. Then it follows that $q, r, s$ are pairwise coprime and $a = pqsa'$, $b = pqrb'$, and $c = prsc'$, with $a', b', c'$ pairwise coprime as well.
Then, $\frac{(pqrsa'b'c')^2}{(pqrsa'b')(pqrsb'c')(pqrsc'a')} = \frac{p^2}{(pq)(pr)(ps)}$.
Let $p = (a, b, c)$, $pq = (a, b)$, $pr = (b, c)$, and $ps = (c, a)$. Then it follows that $q, r, s$ are pairwise coprime and $a = pqsa'$, $b = pqrb'$, and $c = prsc'$, with $a', b', c'$ pairwise coprime as well.
Then, $\frac{(pqrsa'b'c')^2}{(pqrsa'b')(pqrsb'c')(pqrsc'a')} = \frac{p^2}{(pq)(pr)(ps)}$.
"Questions we can't answer are far better than answers we can't question"