How to prove that there exist a concave function and $\gamma\in[0,1]$ and some other numbers which satisfy an inequality I'm working on an economics paper, and in the model I've made I've basically gotten myself a little bit stuck. I need to show that there exists a nondecreasing concave function $u$ and numbers $P$ and $\theta$ with $P>\theta>0$, and $\gamma\in[0,1]$ such that:$$u(P-\theta)-u(-\theta)>\frac{1}{1-\gamma}(u(P)-u(0))$$
And in fact what I would like to show is that for any monotonically increasing $u$ which is strictly concave, we can find $P$, $\theta$, and $\gamma$ which satisfy the equality. I don't have much experience proving that kind of a statement though and I'm having some trouble getting started. Does anyone have any idea what would be a good way to start proving that statement (if it's even true—when I draw pictures of what I want it looks like it should be true but maybe it's not)?
EDIT
I've been puzzling over this and I realize that it will follow directly from a lemma: for $a>b$ and $c>0$, $u(a)-u(b)>u(a+c)-u(b+c)$, since then I can say that $u(P-\theta)-u(-\theta)>u(P)-u(0)$ just by adding $\theta$ to the arguments of $u$ on the left hand side. So I just need to prove that lemma. I can see geometrically why it must be true but I still can't quite make it follow form the concavity of $u$. I'm close though, so I may end up just answering my own question. Writing it out like this is helping.
 A: Consider the function $h(x)=U(x+c)-U(x)$. Its derivative is $U\prime(c+x)-U\prime(x)$. Because $U(x)$ in concave then $U\prime\prime<0$, which means that $U\prime$ is decreasing and hence $h\prime$ is negative. Hence $h(x)$ is decreasing.
Your inequality follows directly: $h(a)<h(b)$ for $a>b$
We can also prove a more general result without resorting to derivatives. Let $F$ be a concave function in an interval and let $x<y\le z<w$ be points in this interval. Then the following holds:$$\frac{F(y)-F(x)}{y-x} \ge \frac{F(w)-F(z)}{w-z}$$
To prove this write $y$ as a linear combination of $x$ and $w$ thus:$$y=\frac{w-y}{w-x}x+\frac{y-x}{w-x}w$$ This implies:
$$
\begin{align*}
F(y) &\ge \frac{w-y}{w-x}F(x)+\frac{y-x}{w-x}F(w) \\
&= F(x) + \frac{y-x}{w-x}(F(w)-F(x))\\
&= F(x) + \frac{y-x}{w-x}(F(w)-F(y)+F(y)+F(x))\\
\end{align*}$$
This is equivalent to:$$\frac{F(y)-F(x)}{y-x}\ge\frac{F(w)-F(y)}{w-y}$$
In the same way we can prove that (note this goes in the other direction):$$\frac{F(w)-F(y)}{w-y}\le\frac{F(w)-F(x)}{w-x}$$
Combining the last two inequalities leads to our result:$$
\frac{F(y)-F(x)}{y-x}\ge\frac{F(z)-F(y)}{z-y}\ge\frac{F(w)-F(z)}{w-z}
$$
