$S \ge \frac{(a+b+c)^2}{2(a+b+c)}$$1 \le a+b+c \le 9 <=>1 \le(a+b+c)^2 \le 81$$2 \le a+b+c \le 18$$Smax<=>(a+b+c)^2max,2(a+b+c)min=>Smax=\frac{81}{2}$$Smin<=>(a+b+c)^2min,2(a+b+c)max=>Smin=\frac{1}{18}$
$S \ge \frac{(a+b+c)^2}{2(a+b+c)}\
ge
\frac
{a+b+c
}{2
}$
từ đây ti
́n
h ra
$min=
1,max=\frac{
9}{
2}$