The set of all values of m for which $mx^2 – 6mx + 5m + 1 > 0$ for all real x is The set of all values of m for which $mx^2 – 6mx + 5m + 1 > 0$ for all real x is?
The answer given is $0<=m<1/4$
My working: $D>=0$
$=> (-6m)^2 -4(m)(5m+1)>=0$
$=> m(4m-1)>=0$
=> Either $m>=1/4$ or $m<=0$
Where am I going wrong?

Option:
$y=m(x^2-6x+5) +1>0;$
0) $m=0$√
1) $m>0$.
A parabola opening upward.
Minimum at:
$y'=m(2x-6)=0;$ $x=3;$
$y_{\min}=m(9-18+5)+1=$
$-4m+1$;
We require: $y_{\min}= -4m+1>0$, or $m<1/4$;
Combining : $0 \le m < 1/4$.
2) Rule out $m<0$ 
(Parabola opening downward )
