\(f\left(1\right)=a+b+c;f\left(5\right)=25a+5b+c\)
\(f\left(1\right)+f\left(5\right)=a+b+c+25a+5a+c=26a+6a+2c=2\left(13a+3a+c\right)>0\)
\(f\left(1\right)=a.\left(1^2\right)+b.1+c=a.b.c\)
\(f\left(5\right)=5^2.a+b.5+c=25a+5b+c\)
\(f\left(1\right)+f\left(5\right)=a+b+c+25a+5b+c\)
\(f\left(1\right)+f\left(5\right)=26a+6b+2c=2.13a+2.3b+2c=2\left(13a+3b+c\right)>0\)