Conditions for an inequation to hold

Question

I am trying to find the domain of a real number n so that $$nx^2+(n-1)x+(n-1)< 0$$ for any $x\ge 0$. I am thinking to solve as: $$n<0$$ and $$\Delta<0$$ Is this enough? "For any $x\ge 0$ "makes me unsure. Is there perhaps some intuition behind this?

Answer 1

The inequation can be written as

$$n(x^2+x+1)<x+1$$

or $$n<\frac{x+1}{x^2+x+1}$$

since $x^2+x+1>0$. put $$f(x)=\frac{x+1}{x^2+x+1}.$$

we have

$$(\forall x\ge 0) \;\; f(x)>0$$ and

$$\lim_{x\to+\infty}f(x)=0$$

thus $$n\le 0$$

Answer 2

I am thinking to solve as: $\quad n<0$

The inequality does not need to be strict, since for $n=0$ the inequality is satisfied for $\forall x \ge 0$:

$$nx^2+(n-1)x+(n-1) \,\Big{|}_{n=0}= -\underbrace{(x+1)}_{\gt 0}< 0$$

If $n \ne 0$ then it is indeed necessary for $n$ to be negative, otherwise the limit at $x \to \infty$ is $+\infty$.

Therefore $n \le 0$ is a necessary condition. It is also a sufficient condition, since for $n \le 0, x \ge 0\,$:

$$nx^2+(n-1)x+(n-1) = \underbrace{n}_{\le 0}\underbrace{(x^2+x+1)}_{\gt 0}-\underbrace{(x+1)}_{\gt 0}< 0$$