Show that if $a,b,c \in \mathbb{R}$ are such that for all $\lambda \in \mathbb{R}$, $a\lambda^²+b\lambda+c\geqslant0$, then $b^2-4ac \leqslant0$. 
Show that if $a,b,c \in \mathbb{R}$ are such that for all $\lambda \in \mathbb{R}$, $a\lambda^²+b\lambda+c\geqslant0$, then $b^2-4ac \leqslant0$.

I'm reading the book Introduction to Topology by Gamelin and Greene and found this problem there. I'm curious why is this on a topology book?
The reasoning I see here is that since $\lambda \in \mathbb{R} \implies b^2-4ac \geqslant 0$ and this would imply that $b^2-4ac=0$ so $\lambda = \frac{-b}{2a}$? I'm wondering if there is some deeper meaning here from topology point of view?
 A: Why you'd find that problem in such a book has already been addressed, so I'll just give an approach to it. Since $a\lambda^2+b\lambda+c$ is asymptotic for sufficiently large $|\lambda|$ to $b\lambda$ if $a=0$ or $a\lambda^2$ otherwise, $a>0$. Then$$a\lambda^2+b\lambda+c=a(\lambda+b/2a)^2-(b^2-4ac)/(4a)$$is non-negative on $\Bbb R$ iff it's non-negative at its minimum (obtained with $\lambda=-b/2a$), i.e. iff $(b^2-4ac)/(4a)\le0$.
A: Looking at the book, this is part 1 of a longer exercise. You are meant to use part 1 to prove part 2, which is known as the Cauchy-Schwarz inequality.
The Cauchy-Schwarz inequality is important in a book about topology, because it can be used to show that Euclidean space is an example of a metric space, an important type of topological space. (The particular chapter you are reading deals with metric spaces.)
As for why this is true: by considering the behaviour as $\lambda \to \infty$, we see that we must have $a\ge 0$. Therefore, after multiplying through by $4a$, we get
$$
4a^2\lambda^2+4ab\lambda+4ac\ge0\,.
$$
Now, completing the square:
$$
(2a\lambda + b)^2  - b^2 + 4ac \ge 0\,,
$$
and so
$$
(2a\lambda+b)^2 \ge b^2 - 4ac
$$
for all $\lambda$.
Now we have to split into two cases. Firstly, if $a=0$, then $b$ must also be $0$, or $a\lambda^2+b\lambda+c=b\lambda+c$ must necessarily take negative values. Otherwise, $2a\lambda + b$ is a linear function with non-zero gradient and therefore intersects the $x$ axis for some value of $\lambda$. It follows that for some $\lambda$ we have $(2a\lambda + b)^2 = 0$ and therefore that $b^2-4ac \le 0$.
A: A geometrical point of view
$a\lambda² + b\lambda +c = 0$ is the equation of a parabola with a vertical axis. The $y$ coordinate of its vertex is $\frac{4ac-b^2}{4a}$. Furthermore, when $a>0$, the parabola points upward, whereas when $a<0$ it points downward.
So when $a>0$, the expression $a\lambda² + b\lambda +c ≥ 0$ means that the whole parabola is above the $x$-axis, thus meaning that the vertex is also above (or on, when $b²=4ac$) the $x$-axis and thus $4ac-b² ≥ 0$, i.e. $b²-4ac ≤ 0$.
When $a<0$, instead, the expression $a\lambda² + b\lambda +c ≥ 0$ cannot hold for each $\lambda \in \mathbb{R}$, but we can state an analogous result:
If $a<0$ and if $a\lambda² + b\lambda +c ≤ 0 \quad \forall \, \lambda \in \mathbb{R}$ (which means that the parabola is pointing upward and is all below the $x$-axis), then $b²-4ac ≤ 0$ (the vertex as well must be below the $x$-axis).
A: Let $$f(\lambda) = a\lambda^2 + b\lambda + c$$We know that if $\Delta = b^2 - 4ac\gt0$ then there are two distinct roots $\lambda_1 \lt \lambda_2$. Furthermore, when $\lambda_1 \lt\lambda \lt \lambda_2$ we have $\text{sign}\{f(\lambda)\} = -\text{sign}\{a\}$ and when $\lambda\gt \lambda_2$ or $\lambda\lt \lambda_1$ we have $\text{sign}\{f(\lambda)\} = \text{sign}\{a\}$. This means $\Delta>0$ always leads to a sign change in $f(\lambda)$ and $\forall \lambda \in \mathbb{R}, \ f(\lambda)\ge0$ can't holds. The only option is $\Delta \le 0$ which leads to no sign change.
A: One more way: Let $a>0$
$f(x)=ax^2+bx+c$, if $f(x) \ge 0, \forall x \in Re$, then $min(f(x))\ge 0$
As $f'(x)=0 \implies x_0=-b/(2a), f''(x_0)=2a>0$ so min is at $x=x_0$, therefore $f(x_0) \ge 0 \implies c-b^2/(4a)\ge0 \implies b^2-4ac\le 0.$
