Looking for simple, continuous differentiable decreasing function example that satisfies several conditions Is there some simple function of $c(s)$
  defined on $s\in[0,1]$
 , which is continuous, differentiable and decreasing in s
  such that $$-\frac{\frac{d\left(c(s)\right)}{ds}}{\frac{c(s)}{s}}>1+\frac{sc(s)}{\int_{s}^{1}c(s)ds}$$
 satisfied at some $s_{0}$
  in [0,1]
 ? 
I studied the question again and find that the above question can be simplified as the following: 
Is there a continuous, decreasing and differentiable function $c(x)\geq0$
  defined on $x\in[0,1]$
 , such that $$\omega(x)=\frac{x\,c(x)}{\int_{x}^{1}c(t)dt}$$
 is decreasing in $x$ at some $x_{0}\in[0,1]$
 ?
This way, we can view $xc(x)$ as a rectangle area below the curve of $c(x)$ and $\int_{x}^{1}c(t)dt$ another area below $c(x)$ curve. The question is exactly when $x$ increase, the rectangle area decrease faster than the integral area. Is there such a curve?
 A: Yes.  We get: 
$$ \omega'(x) = \frac{\left(\int_x^1 c(t)dt\right)[c(x) + xc'(x)] + xc(x)^2}{\left(\int_x^1c(t)dt\right)^2} $$
Suppose we want $\omega'(1/2)<0$.  So we want: 
$$ \left(\int_{1/2}^1 c(t)dt\right)[c(1/2) + (1/2)c'(1/2)] + (1/2)c(1/2)^2 < 0 \quad (Eq *)$$
So just define $c(x)$ to have a very large (negative) derivative at 1/2, but that lasts for a short time only, so the integral $\int_{1/2}^1 c(t)dt$ is still significant. For example fix a small $\delta>0$ and define $c(x)$ so its derivative is: 
$$ c'(x) = \left\{\begin{array}{cc}
-1 & \mbox{if $x \in [0, 1/2-\delta]$}\\
\mbox{tent(x)} & \mbox{if $x \in [1/2-\delta, 1/2+\delta]$} \\
-1 & \mbox{if $x \in [1/2+\delta, 1]$} 
\end{array}\right. $$ 
where $tent(x)$ is a "tent" function such that $tent(1/2-\delta)=tent(1/2+\delta) = -1$ and $tent(1/2) = -M$ for some large value of $M$. You can choose $\delta$ much smaller than $1/M$ so the affect of the tent does not dramatically change the $c$ values for any of the terms in (Eq *) except for $c'(1/2)$.  
For example if $c(0)=10$ and $\delta$ is very small, then $c(x) \approx 10-x$ for all $x \in [0,1]$, whereas $c'(x)=-M$, which is as negative as we like, so (Eq *) is easy to satisfy. 
A: You can pick your $w(x)$ and get your $c(x)$ by solving a differential equation.
Rewrite your expression as
$$\int_x^1 c(t) dt = \frac{1}{w(x)} x c(x)$$
and take a derivative to get
$$-c(x)=(x/w(x)) c'(x) + (x/w(x))' c(x)$$
which can be rewritten as
$$c'(x) + \frac{(x/w(x))'+1}{x/w(x)} c(x) = 0.$$
This is a 1st order linear differential equation and can be solved with an integrating factor.
