Level sets and Concave function in my Calculus textbook about Concave functions the following is derived from this expression:
Let $f$ be a function defined in a convex set which is subset of $\mathbb R^n$ and $Df(x)$ the first derivative gradient of this function.
$f$ is concave iff for every $x,y$ belonging to $U.$
$$f(y)-f(x)< Df(x)(y-x)$$
and from this expression the author somehow concluded that:
Given that the gradient is perpendicular to the tangent plane of the level set of $f(x)$ in a given $x$ if $f(x)$ is concave then all the level set which satisfied ($z$ given that $f(z)\geq f(x)$) are above the tangent plane where above means in the direction of the increasing values of $f$. Also it has this figure:

Can someone explain me how to conclude this from the above definition?
Thanks in advance
 A: EDIT: In my original post I completely misunderstood what was being asked. 
Define the set
$$S = \{x: f(x) \geq \alpha\}$$
where $f \in C^1$ is concave. Can we use the concavity and differentiability of $f$ to know what $S$ looks like without solving the problem analytically?
Let's first try to find an inner box. We locate points $x_1, x_2, \dots, x_p$ for which $f(x_i) = \alpha$. Using the concavity of $f$, the set $S$ must surely contain all elements on the lines between these vertices, i.e. $S$ must contain the inner box defined by the vertices $x_1, x_2, \dots, x_p$. 
It seems (to me) that the original authors wish to point out that we can use the differentiability of the function $f$ to make an outer box of the set using the tangent hyperplanes defined by the gradients at $x_1, x_2, \dots, x_p$. Further if we let $p \rightarrow \infty$ then we have $S$ exactly. 
In the following illustration I show what such an inner box and outer box look like for $p=4$.

I hope this made more sense than my original answer.
A: The missing piece of the puzzle is that for vectors,
$$
cos(\theta) = \dfrac{x \cdot y}{||x|| * ||y||}
$$
where $\theta$ is the angle between $x$ and $y$.  If $x$ and $y$ make an acute angle, then $\dfrac{x \cdot y}{||x|| * ||y||} \ge 0$.
For a concave function, suppose $f(z) \ge f(x)$ like you say.  Then
$$
\dfrac{\nabla f(x) \cdot (z-x)}{||\nabla f(x)||*||z-x||} \ge \dfrac{f(z)-f(x)}{||\nabla f(x)||*||z-x||}\ge 0
$$
so that for any vector $z$ that improves $f$, the gradient and $z-x$ form an acute angle.
More generally, quasi-concave functions have the property that $f$ is quasi-concave iff $f(z)\ge f(x)$ implies $ \nabla f(x) \cdot (z-x) \ge 0$, which is more obviously related to the geometry of the situation. The gradient always makes an acute angle with vectors that improve the objective, and is orthogonal to the supporting hyperplane to the indifference curve.  Concave functions are quasi-concave, since the other definition of quasi-concave is that $f$ is quasi-concave iff $f(\lambda x + (1-\lambda)y) \ge \min\{f(x),f(y)\}$ for all $\lambda \in (0,1)$, which gives a lot of intuition about how the upper contour sets and gradients behave.  It is also an ordinal, and not cardinal property, like concavity.
I like de la Fuente's or Ok's math econ books much better.  Simon and Blume is somehow very encyclopedic without being very precise.
