A nonlinear differential inequality I was trying to prove a generalisation of maximum principle and for that purpose I added a correction term. After some manipulations the condition I was looking for was reduced the following nonlinear differential inequality for some $C^2$ function $x : \mathbb{R} \to \mathbb{R}$
$$x'' + (x')^2 < 0 \qquad \text{in} \quad \mathbb{R}  $$ 
I haven't been able to construct any such $x$. Is the corresponding ODE some standard form? My knowledge of ODEs is very limited so any ideas/hints are welcome.
 A: Note that you are demanding the function $g\colon t\mapsto\exp(x(t))$ to have strictly negative second derivative on the whole of $\mathbb{R}$, in particular, it must be concave and bounded below.  This cannot happen:- the derivative must be nonzero somewhere, WLOG at $t=0$, then the graph lies below the support line at $0$ so must $\to-\infty$ at one of $\pm\infty$.
A: Short version: 1, define $y=x'$; 2, use differential form of Bihari's theorem.
Full version:
\begin{equation}
x''+(x')^2<0\to y'+y^2<0\to \frac{y'}{y^2}+1<0\to \frac{d}{dx}\left(-y^{-1}+x\right)<0,
\end{equation}
then we can conclude
\begin{equation}
-y^{-1}+x<y^{-1}(0)\to y^{-1}>x-y^{-1}(0)\to 1/x'>x-1/x'(0).
\end{equation}
After that we should discus the signs of $1/x'$ and $x-1/x'(0)$,and there will be two cases
\begin{equation}
y>0,\qquad y< \frac{y(0)}{x y(0)-1},
\end{equation}
or
\begin{equation}
x'>0,\qquad x'< \frac{y(0)}{x y(0)-1}.
\end{equation}
The first one is trivial, the second one gives
\begin{equation}
\frac{x y(0)}{1-x y(0)}+x<0.
\end{equation}
