$f^2+(1+f')^2\leq 1 \implies f=0$ 
Find all $f\in C^1(\mathbb R,\mathbb R)$ such that $f^2+(1+f')^2\leq 1$

It's quite likely the answer is $f=0$.
Note that $|f|\leq 1$ and $-2\leq f'\leq 0$.
Therefore $f$ is decreasing and bounded.
What then ? I tried contradiction, without success.
 A: Hints: As you mentioned, $f$ is decreasing and bounded. Think about $\lim_{n \to \infty} f(n)$. Must this limit exist? What does this imply for the limit of the derivative $f'$?
Full Solution. The function $f(x)$ is decreasing and bounded, so $\lim_{x \to \infty} f(x)=L$ for some $L \in [-1,1]$. For the sake of contradiction, we suppose $|L|>0$. To set up the contradiction, we relate $|f(x)|$ and $f'(x)$:
Let $\epsilon\in (0,1]$, and suppose that we have $0 \geq f'(x) \geq -\epsilon$ for some $x \in \mathbb{R}$. Then
\begin{align*}
f^2(x) & \leq 1-(1+f'(x))^2\\
  &\leq -2f'(x) - (f'(x))^2 \\
 &\leq -2f'(x)  \\
 & \leq 2\epsilon.
\end{align*}
Thus $|f(x)| \leq \sqrt{2\epsilon}$. Therefore we know that if $|f(x)| > \sqrt{2\epsilon}$, then $f'(x) <-\epsilon$.
For sufficiently large $x$, we must have $|f(x)| > |L|/2=\sqrt{2(|L|^2/8)}$, hence $f'(x) <-|L|^2/8$. This contradicts the fact that $f(x)$ is bounded below. An entirely analogous argument shows that $\lim_{x \to -\infty} f(x)=0$. Monotonicity implies $f=0$.QED
A: The equation is equivalent to
$$
f^2+2f'+f'^2\le0\tag{1}
$$
Since $f^2+2f'\le0$, where $f\ne0$, we have
$$
(1/f)'\ge\color{#C00000}{1/2}\tag{2}
$$
If $f(x_0)=a\gt0$, then $\dfrac1f(x_0)=\dfrac1a\gt0$ and $(2)$ says that
$$
\frac1f\left(x_0-\frac3a\right)\le\frac1f(x_0)-\color{#C00000}{\frac12}\frac3a\lt0\tag{3}
$$
as long as $\dfrac1f$ doesn't pass to $-\infty$ in $\left[x_0-\frac3a,x_0\right]$.
In any case, on $\left[x_0-\frac3a,x_0\right]$, $\dfrac1f$ must pass through $0$, which is impossible because $f\in C^1(\mathbb{R})$.
If $f(x_0)=a\lt0$, then $\dfrac1f(x_0)=\dfrac1a\lt0$ and $(2)$ says that
$$
\frac1f\left(x_0-\frac3a\right)\ge\frac1f(x_0)-\color{#C00000}{\frac12}\frac3a\gt0\tag{4}
$$
as long as $\dfrac1f$ doesn't pass to $\infty$ in $\left[x_0,x_0-\frac3a\right]$.
In any case, on $\left[x_0,x_0-\frac3a\right]$, $\dfrac1f$ must pass through $0$, which is impossible because $f\in C^1(\mathbb{R})$.
Therefore, $f(x)=0$ for all $x\in\mathbb{R}$.
A: Since $f(x)$ is bounded and decreasing both $\lim_{x \rightarrow \infty} f(x)$ and $\lim_{x \rightarrow -\infty} f(x) $ exist. If $f(x)$ were not identically zero, then at least one of these limits is nonzero. Say it is the first one, and call the limit $L$. 
By the mean value theorem, $f(n+1) - f(n) = f'(x_n)$ for some $x_n$ between $n$ and $n + 1$. The left-hand side of this equation converges to $L - L = 0$ as $n$ goes to infinity, so we have
$$\lim_{n \rightarrow \infty} f'(x_n) = 0$$
But we also have
$$\lim_{n \rightarrow \infty} f(x_n) = L$$
Plugging $x_n$ into $f(x)^2 + (1 + f'(x))^2 \leq 1$ and taking limits as $n$ goes to infinity gives $L^2 \leq 0$, a contradiction. 
A similar argument works if $\lim_{x \rightarrow -\infty} f(x) \neq 0$.
A: I think I have to completely rewrite the solution keeping the wrong one above untouched. As I said before I am sure we can build such function, and I think I did it using the ODE in the above solution in the end. 
I am building a counterexample function:
Fistly, $f(x)= 0$ for $x\le 0$;
Now I'd like to build a simple function satisfying condition 
$$
f^2+(1+f^\prime)^2\le 1
$$ 
on some interval $[0,x^*]$.
I define 
$$
g(t)=-x^3/3-x^2/2\\
g^\prime(x)=-x^2-x
$$ 
I is obvious that in some positive neighborhood $(0,\epsilon)$
$$
g^2+(1+g^\prime)^2 =O(\epsilon^4)+1-O(\epsilon) \le 1
$$
It is obvious also that staring from some $x^*$ 
$$
g^2+(1+g^\prime)^2\ge 1
$$
Besides we have this point $x^*$ is such that 
$$
g^2(x^*)+(1+g^\prime(x^*))^2= 1
$$
Let's check condition $g^\prime(x^*)>-1$. It is easy t estimate that $x^*$ is about $0.9$ and that then $g^\prime(x^*)< -1
Now again consider the differential equations
$$
f^\prime=-1+\sqrt{1-f^2}\\
f^\prime=-1-\sqrt{1-f^2}\\
$$
and choose the second one  in accordance with the sign $g^\prime(x^*)+1$.
The solution of this equation with initial condition $f(x^*)=g(x^*)$ will extend our function on $R$. 
So the final function $f(x)$ is 
$$
f(x)=0 ~ if ~ x\le 0 \\
f(x)=g(x) ~ if~ 0<x\le x^* \\
solution~ of~ f^\prime=-1-\sqrt{1-f^2}, f(x^*)=g(x^*) ,~ x\ge x*\\
$$
Since I did not find any mistake in the proof that such function cannot exist I again probably made mistake somewhere. But I cannot find it. Any comments please.
May be this ODE cannot have a solution on $R$? Lipschitz condition is not satisfied. 
A: FIX
Well the solution of differential equation  with non zero initial condition will satisfy your property
$$
f'=-1-\sqrt{1-f^2}
$$
Edit
Notice that equation 
$$
f'=-1+\sqrt{1-f^2}
$$
is also OK. 
Now let's change $\tau = -t$ and rewrite the second equation as function of $\tau$
$$
f_\tau^\prime =1-\sqrt{1-f^2}
$$
Let's build now the function on $R^+$ as a solution of the first eqaution and on $R^-$ as a solution of the third equation. To guarantee differentiability in $t=0$
let's make equal derivatives at time $t=\tau=0$.
$$
f^\prime_t=-1-\sqrt{1-f^2}=-f^\prime_\tau=-1+\sqrt{1-f^2}
$$
So with initial condition $f=1$ we can propagate this function on $R$.
I changed signs here. Now the initial derivative is  
