If $ax^2+bx+c<x$, for any $x$ then is $b\ge 1$ or $b <1$ or $c=0$ or $a\le 0$?

Question

I tried by using the following : $ax^2+bx+c<x$ $\implies (b-1)^2<4ac$ But $ax^2+bx+c<x$ also $\implies$ a parabola lying underneath a straight line $y=x$ such that the parabola faces the origin from negative y axis. In that case is $b<1$ the correct choice?

Answer 1

$ax^2+{(b-1)}x + c < 0$ for all $x$ when $a\neq 0$ implies both

(1) $a < 0$

(2) $(b-1)^2-4ac < 0$

For $a=0$, this becomes a linear inequality $(b-1)x+c<0$ and is true for all $x$ when $b=1$ and $c<0$

Answer 2

Consider the function

$$ax^2+(b-1)x+c<0.$$

If $a\ne0$, it is a parabola. To stay in the negatives, it needs to be concave down, i.e. $a<0$. Then it may not have real roots, $(b-1)^2<4ac$ (implicitly, $c<0$).

If $a=0$, the function is linear and takes both signs, unless $b=1$. In this case, only $c<0$ remains.

The complete condition is

$$(a<0\land (b-1)^2<4ac)\lor (a=0\land b=1\land c<0).$$

Only $a\le0$ is compatible.