$f^2(x)+(f^\prime(x))^2 \geq 1$ $f(0)=1$ implies $f^\prime(t)=0$ Let $f:\mathbb{R}\rightarrow [0,1]$ be continuously differentiable function satisfying:
$f^2(x)+(f^\prime(x))^2 \geq 1$ $\forall$ $x\in\mathbb{R}$ and $f(0)=1$.
Prove that there exists $t\gt0$ such that $f^\prime(t)=0$
Here we can see that the function attains its maximum at $x=0$. So $f^\prime(0)=0$.We have to prove that for some $t\gt0\quad f^\prime(t)=0$. Suppose that no such t exists. Then $f^\prime(t)$ must be less than $0$ for all $t\gt0$. Then f will be decreasing for $t\gt0$. This is what I have done so far. I can see that this is not possible (graphically) because such a function will violate the given condition $f^2(x)+(f^\prime(x))^2 \geq 1$ for all $x\in\mathbb{R}$. But I can not prove it theoretically.
How can we proceed from here? Thanks in advance. 
 A: Suppose that $f'(t)\neq 0$ to reach a contradiction. 
Since $f(0)=1$ and $f(t)\leq 1$ for all $t$, we must have that $f'(t)<0$ for all $t>0$.
Manipulating the expression we find that 
$$-f'(x)\geq \sqrt{1-f^{2}(x)    }.$$
Now by integrating we find 
$$f(0)-f(t)\geq \int _{0}^t\sqrt{1-f^{2}(x)    } \rm{d}x$$
Since $f$ decreasing $\lim_{t \to \infty} f(t)=c\geq 0$. Again since $f$ is decreasing we can find a $x_0$ such that $1>c'>f(x)\geq 0$ for $x\geq x_0$.
We take now $t \to \infty$
\begin{align}1-c&\geq \int _{0}^{\infty}\sqrt{1-f^{2}(x)    } \rm{d}x\\
&=\int _{0}^{x_0}\sqrt{1-f^{2}(x)    } \rm{d}x+\int _{x_0}^{\infty}\sqrt{1-f^{2}(x)    } \rm{d}x\\
&>\int _{0}^{x_0}\sqrt{1-f^{2}(x)    } \rm{d}x+\int _{x_0}^{\infty}\sqrt{1-c'  ^2  } \rm{d}x=\infty
\end{align}
A contradiction since $1-c \leq 1-0<\infty$.
A: Suppose $f'$ is always negative on $(0, \infty)$; then $f$ is decreasing and bounded below by 0; let $\lim_{x\to \infty} f(x) = M \in [0, 1)$. Pick $N> 0$ such that $f(x) \in [M, (1+M)/2]$ for all $x \geq N$; for all $T>0$, let $x_T \in [N, N+T]$ such that $$f'(x_T) = \frac{f(N+T) - f(N)}{T}$$
Then $|f'(x_T)| < \frac{1}{T} \frac{1-M}{2}$ and $|f(x_T)| < \frac{1+M}{2} <1$. Now let $T \to \infty$ to obtain a contradiction.
A: It suffices to consider the case when, for $x>0$, $f$ is strictly decreasing and never obtains the value $0$. Thus the derivative $f'<0$ cannot be smaller than $-1/m$ for any $m\in\mathbb{N}_+$ for too long (after $x>0$). It follows that, for any $m$, one can find an interval $I_m:=(a_m,b_m)$ where $0 < a_m < b_m$ such that in $I_m$ we have $f'(x) > -1/m$. Therefore $(f'(x))^2$ can too be made arbitrarily close to $0$ by choosing an appropriate $I_m$.
But we must also guarantee $(f(x))^2+(f'(x))^2 \geq 1$ everywhere, and yet for any $m\in\mathbb{N}_+$ we have $(f(x))^2 \geq1-(f'(x))^2 > 1-1/m$ in the interval $I_m$. So $f(x)$ can be made to be arbitrarily close to $1$ by picking an appropriate $I_m \subseteq (0,+\infty)$. Impossible.
