How to prove the relationship of two simple functions? Given two functions $f(x)>0$ and $g(x)>0$ on $x\in[a,b]$, if they satisfy the following constraints:
\begin{cases}
f''(x) >0 \\
g''(x) = 0\\
f(a)=g(a)\\
\int_a^bf(x)\mathrm{d}x= \int_a^bg(x)\mathrm{d}x
\end{cases},
then the following conclusions hold: 


*

*$f(x)$ and $g(x)$ must have and only have one intersection on interval $(a,b)$. 

*If we denote the x-coordinate of the intersection with $s$, then: 
\begin{cases}
f(x)<g(x), \qquad \mathrm{if}\quad x \in (a,s);\\
f(x)>g(x),\qquad \mathrm{if}\quad x \in (s,b)
\end{cases}


But how to prove it?
Can you help me?

Here's an idea. Notice that if there is never a point in the interval $[a,b]$ at which $f(x)>g(x)$, then the area under both curves could not be equal.
Furthermore, since these two curves have the same area under the curves in the interval $[a,b]$, it should be obvious that 
$\int_a^s g(x)-f(x)dx=\int_s^bf(x)-g(x)dx$
This means that portion of $g(x)$ above $f(x)$ is equal to the portion of $f(x)$ above $g(x)$ on the interval $[a,b]$ which holds because $f(x)$ has to "make up for" the area it was under $g(x)$.
