Prove that there exists at least one $x_0\in\mathbb{R}$, such that $f(x_0)+f''(x_0)=0$ 
Let $f:\mathbb{R}\to\mathbb{R}$ be a function, two times
  differentiable with $\left|f(x)\right|\leq1,\forall x\in\mathbb{R}$
  and $f^2(0)+\left(f'(0)\right)^2=4$. Prove that there exists at least
  one $x_0\in\mathbb{R}$, such that $f(x_0)+f''(x_0)=0$.

I have tried the usual Rolle method by $e^x$, but didn't go far... Any help available?
 A: Let's define $g(x) = f(x)^2 + f'(x)^2$. We have:


*

*$g'(x) = 2f'(x)(f(x)+f''(x))$

*$g(0) = 4$


We will find a local maximum of $g$, noted $x_0$, which verifies $g(x_0) \ge 4$. We shall distinguish two options:


*

*$g(0) = 4 \ge g(x) \ \forall \ x \in \mathbb{R}$, in which case $0$ is the local maximum that we are trying to find.

*There exists $y \in \mathbb{R} \setminus \{0\}$ with $g(y) > 4$. Let's supose that $y > 0$ (otherwise apply the following reasoning to $-g$) and try to find a number $x' > y$ which verifies $g(x') = 4$. If there isn't such a number, then we would have $g(x) > 4$ for every $x > y$. Thus, $\left| f'(x) \right| > \sqrt 3$ for every $x > y$. Using the mean value theorem on $x > \max\{2,y\}$ we get a contradiction:
$$ 2 \ge \left| f(x) - f(0)\right| = \left| f'(\xi_x) \ x \right| > 2 \sqrt 3 > 2 $$
As a consequence, we can take $x' > y$ with $g(x') = 4$. The function $g$ has a global maximum in $[0, x']$ because it is continuous. Let $x_0$ be that global maximum. Then, $g(x_0) \ge g(y) > 4$ and, consequently, $x_0 \ne 0, x'$. Hence, $x_0$ is a local maximum of $g$ with $g(x_0) \ge 4$. 
Finally, since $x_0$ is a local extrema we have: 
$$ 0 = g'(x_0) = 2f'(x_0)(f(x_0)+f''(x_0)) $$
Furthermore, $f'(x_0)^2 \ge 4 - f(x_0)^2 \ge 3 > 0$ and, thus, $f(x_0)+f''(x_0) = 0$.
A: Try to explore the extrema of 
$$
g(x)=f(x)^2+f'(x)^2\implies g'(x)=2f'(x)(f(x)+f''(x)).
$$
