Áp dụng BĐT Mincopxki ta có:
$S\ge\sqrt{(a+b+c)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}$
$\ge\sqrt{(a+b+c)^2+\dfrac{81}{(a+b+c)^2}}$
$\ge\sqrt{(a+b+c)^2+\dfrac{81}{16(a+b+c)^2}+\dfrac{1215}{16(a+b+c)^2}}$
$\ge\sqrt{\dfrac{9}{2}+\dfrac{1215}{16.\frac{9}{4}}}=\dfrac{3\sqrt{17}}{2}$
$\min S=\dfrac{3\sqrt{17}}{2} \Leftrightarrow a=b=c=\dfrac{1}{2}$