Prove that: $(f(x))^{2} + (f'(x))^{2} \leqslant \max(a,b)$ where $(a,b) \in \mathbb{R}^2$ Let $f\in C^2(\mathbb{R},\mathbb{R})$. 
Assume there exist  $(a,b) \in \mathbb{R}^2$ such that  $\forall x \in \mathbb{R}, (f(x))^{2} \leqslant a$ and $(f'(x))^{2} + (f''(x))^{2} \leqslant b$. 
Prove that:
$$(f(x))^{2} + (f'(x))^{2} \leqslant \max(a,b)$$ for all $x \in \mathbb{R}$.


*

*I tried to use taylor lagrange with the integral form of the remainder but I did not succeed. 
Ì really don't know how to tackle  the problem.


Any suggestion, Hint (or Answer) will be very helpful,
Thank you in advance,
 A: For typographical reasons, I'm going to switch to Newton-style $\dot f$ notation for derivatives.  
It is natural to consider the function $F=f^2+\dot f^2$ and its derivative $\dot F = 2\dot f (f+\ddot f)$. At any point of extremum of $F$ we have either 


*

*$\dot f=0$ — hence $F=f^2\le a$, or

*$f+\ddot f=0$ — hence $F= \ddot f^2+\dot f^2 \le b$


Unfortunately this isn't enough, because it is conceivable that $F$ could grow at infinity without having any points of extremum. We have to work harder.
Suppose the conclusion is false. Then there is $\epsilon>0$ such that the set
$$U=\{f^2+\dot f^2> \max(a,b)+\epsilon\}$$ is nonempty. Being open, $U$ can be written as the disjoint union of open intervals. Let $I$ be  such an interval. At every point of $I$ we have $\dot f^2>\epsilon$. By the mean value theorem, 
$$|f(u)-f(v)|\ge \sqrt{\epsilon}\,|u-v|,\qquad \forall u,v\in I$$
Since $f$ is bounded, it follows that $I$ is a finite interval. By its construction, $F=\max(a,b)+\epsilon$  at the endpoints of $I$. Therefore, $F$ attains a maximum inside of $I$, to which the reasoning from the beginning of the post applies.
