Proof from Calculus 1 Last days, from going into a website of the university of Pisa, I found an exercise given in the previous exams, in 1999.
The problem was like:

Given a continuous function $f$ in $\mathbb R$, and which satisfies the relation:
     $$4\left(f(x)\right)^2   -   4 f(x) + 1 > 0, \forall x \in \mathbb R$$
     show that if a point $x_0$ exists such that $f(x_0) = 0,$ then the function has an   upper limit . .

I tried to solve that, and think about it, but didn't know how to do it. What could a possible proof be?

From the given relation, we have equivalently that:
$$[2f(x) - 1]^2 >0.$$
Thus, we know that there is no point $x_1$ such that $f(x_1) = \frac 12.$
But we know that  $f$ is continuous in $\mathbb R$. Thus, the following proposition holds.
$$\quad \text{Either } f(x) < \frac 12,\,\forall x\in\mathbb R \text{ or  } f(x)>\frac 12,\,\forall x\in\mathbb R.^*$$
Since there is a $x_0$ such that $f(x_0) = 0,$ we are in the first case, thus $\frac 12$ is an upper limit of $f$.

*Proof of the proposition: Suppose that there is a $\xi_1$, such that $f(\xi_1)<\frac 12 $ and a $\xi_2$, such that $f(\xi_2)>\frac 12.$ Then, by applying the intermediate value theorem on $[\xi_1,\xi_2]$, we would have that there is a $\xi_0 \in (\xi_1,\xi_2),$ such that $f(\xi_0) = \frac 12,$ which cannot hold. 
