\(\dfrac{r}{R}=\dfrac{\dfrac{S}{p}}{\dfrac{abc}{4S}}=\dfrac{4S^2}{abc.p}=\dfrac{4\left(p-a\right)\left(p-b\right)\left(p-c\right).p}{abc.p}\\ =\dfrac{\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)}{2.abc}\left(...\right)\)
mà
\(\sqrt{b+c-a}.\sqrt{a+c-b}\le\dfrac{b+c-a+a+c-b}{2}=c\)
tương tự .........
\(\Rightarrow\left(...\right)\le\dfrac{1}{2}\)