Prove: if the sum of squares of a bounded function and its derivative is greater than $1$ then the derivative is $0$ at some point Let $f : \mathbb{R} → [0, 1]$ be a continuously differentiable function satisfying
$(f(x))^2 + f'^2(x) \geq 1$ for all $x \in \mathbb{R}$ and $f(0) = 1$. Prove that there exists
$t > 0$ such that $f'(t) = 0$.
If $(f(x))^2 + f'^2(x) = 1$ then by integrating I get a periodic function ($\cos x$) and the proof is done.But I have no Idea how to go further when it is greater than 1.
 A: This is a not fully rigorous proof. By contradiction, let's assume that $\forall t, f(t) \neq 0$
Because $f(0) = 1$ and $f \leq 1$, necessarily $f' < 0$ on a small interval $]0,\delta[$. Thus, $f(t) < 1-\epsilon$ when $t > \delta$. By assumption : $$ (1-\epsilon)^2 + f'(t)^2 \geq f(t)^2 + f'(t)^2 \geq 1.$$ Consequently, $ f'(t) \leq - \sqrt{1 - (1- {\epsilon})^2} \neq 0$ for all $t > \delta$ which contradicts $f \geq 0$.
A: If $0\leq f(x)\leq 1$ and $f(0)=1$ then $0$ is a global maximum and $f'(0)=0$ by Fermat's lemma. If $f'(x)\neq0$ for $x>0$ then $f'$ has constant sign there, and that sign has to be negative because it can not strictly increase up from $1$. So it strictly decreases, but can not go below $0$, and therefore has a limit $1>b\geq0$. Let $1>b_1>b$ so that $f(x)\leq b_1$ for $x>R$. Then 
$$f'(x) \leq -\sqrt{1 - f^2(x)}\leq-\sqrt{1 - b_1^2}$$
and $f$ has to eventually decrease faster than a linear function with no limit, contradiction. 
By the way, $f(x)= \cos x$ is not the only solution to $f'(x) = \sqrt{1 - f^2(x)}$ with $f(0)=1$, so is $f(x)=1$. The square root is non-Lipschitz at $0$, so there is no uniqueness of solutions. The first function decreases after $0$ and then reverses, and the second stays constant, and this is the best functions under these assumptions can do.
