r > \(\frac{abc}{bc + ac + ab}\)
r < abc
r > \(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{1}{c}\)
\(\frac{1}{abc}\)
Correct answer is A
\(\frac{r}{a}\) + \(\frac{r}{b}\) + \(\frac{r}{c}\) > 1 = \(\frac{bcr + acr + abr}{abc}\) > 1
r(bc + ac + ba > abc) = r > \(\frac{abc}{bc + ac + ab}\)