An Inequality in Calculus 
Let $f(0)>0$ and $f(x)$ increases on $[0,1]$. There exists a positive number $s$ such that
  $$\int_0^1xf(x)\mathrm{d}x=s\int_0^1f(x)\mathrm{d}x$$
  Prove that
  $$\int_0^sf(x)\mathrm{d}x\le\int_s^1f(x)\mathrm{d}x$$

This is a problem from my homework. Let $\displaystyle{g(x)=\int_0^xf(t)\mathrm{d}t},x\in[0,1]$. If $f$ IS CONTINUOUS, we have $g(0)=0, g(1)=1$(WLOG, we can presume that) and $s=\displaystyle{\int_0^1xg'(x)\mathrm{d}x=1-\int_0^1g(x)\mathrm{d}x}$, which is the same to $\displaystyle{\int_0^1g(x)\mathrm{d}x=1-s}$.
If $g(s)>\frac{1}{2}$ and $f$ IS CONTINUOUS, we have $g(x)$ is convex on $[0,1]$. I used some inequalities to show that 
$$\int_0^1g(x)\ge\frac{1-s}{2(1-g(s))}$$ and we get $g(s)\le\frac{1}{2}$, follws the contradiction.
But the question is that I want to find a solution WITHOUT continuity (or this solution can be transformed) and I'd like this progress to be more "analytical" because I established a coordinate system to prove the inequality. Thanks for your help!

Let $g(x)= x$. Clearly, since $0<g(x)<1$ on the interval of integration, $g$ is continuous and $f>0$, we have
$$ 0 = \min_{x \in [0,1]} g(x) \int_0^1 f  < \int_0^1 g(x) f(x) \, dx < \max_{x \in [0,1]} g(x) \int_0^1 f = \int_0^1 f. $$ But therefore
$$ 0 < \frac{1}{\int_0^1 f} \int_0^1 g(x)f(x) \, dx < 1, $$
and by applying the intermediate value theorem to $g$, there is $s \in (0,1)$ so that
$$ s = g(s) = \frac{1}{\int_0^1 f} \int_0^1 g(x)f(x) \, dx, $$
as required (to clarify, the IVT states that $g$ takes on every value between its minimum $0$ and its maximum $1$ at least once in the interval, so we can certainly find one for which this holds).
(Note that for this part, all we needed was that $f$ is positive.)

The following is surely not the best way to do the second part, but does work. We have
$$ \frac{1}{s} \int_0^s f \leq \max_{[0,s]} f \leq f(s) \leq \min_{[s,1]} f \leq \frac{1}{1-s} \int_s^1 f, $$
and we are finished if we can show $s/(s-1) \geq 1$, or in other words, $s\geq 1/2$. This is also easy to check: $\int_0^1 (y-x) f(x) \, dx$ is an increasing function with one zero, at $x=s$, but
$$ \int_0^1 (1/2-x) f(x) \, dx = \int_0^{1/2} (1/2-x)f(x) \, dx + \int_0^{1/2} (1/2-(1-x'))f(1-x')) \, dx' = \int_0^{1/2} (1/2-x)(f(x)-f(1-x)) \, dx \leq 0, $$
since $f(1-x)\geq f(x)$ for $1/2<x<1$. The result follows.
