$\exists c>0$ such that $ (z-x)\int_z^y{f(t)dt} - (y-z)\int_x^z{f(t)dt \geq c(z-x)(y-z)}$ 
Let $f:[a,b] \to \mathbb{R}$ be a strictly increasing function. Prove that, $\forall x, y \in [a,b], x \leq y, \exists c > 0$ such that $$ \displaystyle{(z-x)\int_z^y{f(t)dt} - (y-z)\int_x^z{f(t)dt \geq c(z-x)(y-z)}},$$ $ \forall z \in [x,y]. $

The inequality is equivalent with $ \displaystyle{\frac{\int_z^y{f(t)dt}}{y-z} - \frac{\int_x^z{f(t)dt}}{z-x} \geq c, \forall z \in [x,y]. }$
My assumption is that $\displaystyle{c = \inf_{z \in [x,y]} {\frac{\int_z^y{f(t)dt}}{y-z} - \frac{\int_x^z{f(t)dt}}{z-x}}}$, but I don't know how to prove that this infimum is not $0$. 
I also tried using Riemann sums for the intervals $[x,z]$ and $[z,y]$, but I didn't manage to solve the problem.
 A: The desired inequality
$$
(z-x)\int_z^y f(t) \, dt - (y-z)\int_x^z f(t)\, dt \geq c(z-x)(y-z)
$$
is trivially satisfied for $z=x$ and for $z=y$ with arbitrary $c$
(both sides are zero). Therefore it suffices to show that the function
$h: (x, y) \to \Bbb R$ defined as
$$
h(z) = \frac{\int_z^y f(t) \, dt}{y-z} - \frac{\int_x^z f(t)dt  \, }{z-x} 
$$ 
has a strictly positive lower bound.
Remark: It is clear (from the monotony) that
$$
\frac{\int_z^y f(t) \, dt}{y-z} \ge f(z) \ge \frac{\int_x^z f(t)dt  \, }{z-x}
$$ 
so that $h(z) \ge 0$. The idea of the following proof is to split
the integrals in two parts in order to get a better estimate for the difference.
Let $z \in (x, y)$ and set $w = \frac{z+3y}{4}$. Then
$$
 \int_z^y f(t) \, dt = \int_z^w f(t) \, dt + \int_w^y f(t) \, dt
 \ge (w-z) f(z) + (y-w) f(w) \\
\Longrightarrow 
\frac{\int_z^y f(t) \, dt}{y-z} \ge \frac{w-z}{y-z}f(z) + \frac{y-w}{y-z}f(w) = \frac 34 f(z) + \frac 14 f(\frac{z+3y}{4}) \, .
$$
In the same way (using $v = \frac{3x+z}{4}$ as intermediate point) it can be shown that
$$
\frac{\int_x^z f(t)dt  \, }{z-x} \le \frac 34 f(z)  + \frac 14 f(\frac{3x+z}{4}) \, .
$$
It follows that
$$
h(z) \ge \frac 14 \left( f(\frac{z+3y}{4}) - f(\frac{3x+z}{4}) \right) \\
\ge \frac 14 \left( f(\frac{x+3y}{4}) - f(\frac{3x+y}{4}) \right) =: c > 0
$$
for all $z \in (x, y)$.

Alternative approach: With the substitutions
$t = (1-s)z + sy$ and $t = (1-s)z + sx$, respectively, we get
$$
(z-x)\int_z^y f(t) \, dt - (y-z)\int_x^z f(t)\, dt  \\
= (z-x)(y-z) \int_0^1 \bigl( f((1-s)z+sy) - f((1-s)z + sx) \bigr) \, ds
$$
and it remains to estimate the integral on the right-hand side:
$$
\int_0^1 \bigl( f((1-s)z+sy) - f((1-s)z + sx) \bigr) \, ds \\
 \ge \int_{1/2}^1 \bigl( f((1-s)z+sy) - f((1-s)z + sx) \bigr) \, ds \\
 \ge \int_{1/2}^1 \bigl( f((1-s)x+sy) - f((1-s)y + sx) \bigr) \, ds
 =: c \, .
$$
$c$ is positive because the last integrand is positive and
strictly increasing for $\frac 12 < s \le 1$.
