I have the set
$$ C_c = \{(x,y,z) \epsilon \mathbb{R}^3 : (2x-x^2+y)(2y-3z)(5x-z) > 1, |x| < 1, y > 3, z < 2\} $$
and I need to prove whether it's convex or not. I know that the intersection of different convex sets is convex, but the first inequality is not a convex set (from plotting it). However, the whole set seems to be. What basically makes it convex is the restriction on the values of y (I think). But I don't know how to go about proving that it is convex.
Let's focus on that first inequality. We note four things:
First: all three terms are concave functions of $x$, $y$, and $z$. The second and third terms are affine, of course, but it is their concavity we need here.
Second: the first two terms are positive when the other inequalities are satisfied. For the first term, it helps to note that $2x-x^2+y=y+1-(x-1)^2$. The second term is relatively simple to verify as well.
Third: the third term must be positive as well. Unlike the first two terms, this does not immediately follow from the other inequalities. But since the first two terms of the product are positive, the only way the inequality can be satisfied is if the third term is positive as well.
Fourth: the inequality is equivalent to $$\sqrt[3]{(2x-x^2+y)(2y-3z)(5x-z)}>1.$$ The left-hand site is now a geometric mean of concave, positive terms. The geometric mean is a concave, non-decreasing function of positive arguments. Nonlinear composition rules for convex optimization (see, for instance, Section 3.2.4 of Boyd & Vandenberghe's Convex Optimization) tell us that the composition of a concave, nondecreasing outer function and concave inner functions is concave. So the entire left-hand side is a concave function of $x$, $y$, and $z$ whenever those terms are positive.
Therefore, this product inequality takes the form $f(x,y,z)>1$, where $f$ is a concave function of $x,y,z$. This describes a concave set.
If we need a bit more rigor, we can define the following extended-valued function: $$f(x,y,z)=\begin{cases} \sqrt[3]{(2x-x^2+y)(2y-3z)(5x-z)} & |x|<1,~y>3,~z<2,~5x-z>0 \\ -\infty & \text{otherwise} \end{cases}$$ This function has a convex domain, and is concave on that domain, for all of the reasons we have argued above. Then $C_c$ is described by a single inequality involving this concave function: $$C_c=\{(x,y,z)\,|\,f(x,y,z)>1\}$$
