A linear equation $ax+by+cz=d$ determines a plane $$\pi:=\bigl\{(x,y,z\in{\mathbb R}^3\bigm|ax+by+cz=d\bigr\}$$ in our $(x,y,z)$-space, and the linear inequality $ax+by+cz<d$ one of the two half-spaces defined by this plane $\pi$. You are given $10$ such inequalities, and are told to draw the set $S$ of points that fulfill all of them simultaneously. This set $S$ then is the intersection of ten half-spaces. This intersection is a certain convex polytope, it could also be empty.
The inequalities $0<x<1$, $\>0<y<1$, $\>0<z<1$ obviously define the open unit cube $C:=\>]0,1[\>\times\>]0,1[\>\times]0,1[\>$. Hence we know that $S\subset C$, and a nice drawing should be possible. We then are given the additional inequalities
$$y<x,\qquad z>{x+y\over2},\qquad z>{x\over2}\ ,$$
the first of which is a rewriting of $y<{x+y\over2}$. We already know that $y>0$ in $S$. It follows that the inequality $z>{x\over2}$ is automatically fulfilled when $z>{x+y\over2}$.
In all, we have to draw the part of $C$ where $y<x$ and $z>{x+y\over2}$. The equality $y=x$ splits $C$ diagonally in two halves, and we are left with a triangular prism $P$ standing on the $(x,y)$-plane. This prism has three vertical edges of length $1$. The last plane $\pi:\ z={x+y\over2}$ intersects these edges in the points $(0,0,0)$, $\bigl(1,0,{1\over2}\bigr)$, and $(1,1,1)$. The set $S$ then consists of all points in $P\subset C$ lying above $\pi$.
