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Not all simple polytopes are incribable, e.g. the dual of the cyclic polytope $C_4(8)$ is simple and not inscribable, as shown recently in Combinatorial Inscribability Obstructions for Higher-Dimensional Polytopes by Doolittle, Labbé, Lange, Sinn, Spreer and Ziegler

In dimension $3$, there is a combinatorial criterion by Rivin describing inscribabilty completely. I think already a cube with corner cut, which is simple, will be a non-inscribable $3$-polytope. This can be checked with the following two lines of sage:

sage: C = polytopes.cube().intersection(Polyhedron(ieqs = [[15/8,1,1,1]])) 
....: C.graph().is_inscribable()                                                
False

sage: C.is_simple()                                                             
True

It's nice that Rivin's criterion is implemented in sage...

Here's an image of the graph of the "cube without one corner" 3-polytope, which is non-inscribable and simple:

Not all simple polytopes are incribable, e.g. the dual of the cyclic polytope $C_4(8)$ is simple and not inscribable, as shown recently in Combinatorial Inscribability Obstructions for Higher-Dimensional Polytopes by Doolittle, Labbé, Lange, Sinn, Spreer and Ziegler

In dimension $3$, there is a combinatorial criterion by Rivin describing inscribabilty completely. I think already a cube with corner cut, which is simple, will be a non-inscribable $3$-polytope. This can be checked with the following two lines of sage:

sage: C = polytopes.cube().intersection(Polyhedron(ieqs = [[15/8,1,1,1]])) 
....: C.graph().is_inscribable()                                                
False

sage: C.is_simple()                                                             
True

It's nice that Rivin's criterion is implemented in sage...

Here's an image of the graph of the "cube without one corner" 3-polytope, which is non-inscribable and simple:

I just checked that this is the smallest non-inscribable simple 3-polytope: all other simple 3-polytopes with up to 10 vertices are inscribable.

Not all simple polytopes are incribable, e.g. the dual of the cyclic polytope $C_4(8)$ is simple and not inscribable, as shown recently in Combinatorial Inscribability Obstructions for Higher-Dimensional Polytopes by Doolittle, Labbé, Lange, Sinn, Spreer and Ziegler

In dimension $3$, there is a combinatorial criterion by Rivin describing inscribabilty completely. I think already a cube with corner cut, which is simple, will be a non-inscribable $3$-polytope. This can be checked with the following two lines of sage:

sage: C = polytopes.cube().intersection(Polyhedron(ieqs = [[15/8,1,1,1]])) 
....: C.graph().is_inscribable()                                                
False

sage: C.is_simple()                                                             
True

It's nice that Rivin's criterion is implemented in sage...

Not all simple polytopes are incribable, e.g. the dual of the cyclic polytope $C_4(8)$ is simple and not inscribable, as shown recently in Combinatorial Inscribability Obstructions for Higher-Dimensional Polytopes by Doolittle, Labbé, Lange, Sinn, Spreer and Ziegler

In dimension $3$, there is a combinatorial criterion by Rivin describing inscribabilty completely. I think already a cube with corner cut, which is simple, will be a non-inscribable $3$-polytope. This can be checked with the following two lines of sage:

sage: C = polytopes.cube().intersection(Polyhedron(ieqs = [[15/8,1,1,1]])) 
....: C.graph().is_inscribable()                                                
False

sage: C.is_simple()                                                             
True

It's nice that Rivin's criterion is implemented in sage...

Here's an image of the graph of the "cube without one corner" 3-polytope, which is non-inscribable and simple:

Not all simple polytopes are incribable, e.g. the dual of the cyclic polytope $C_4(8)$ is simple and not inscribable, as shown recently in Combinatorial Inscribability Obstructions for Higher-Dimensional Polytopes by Doolittle, Labbé, Lange, Sinn, Spreer and Ziegler

In dimension $3$, there is a combinatorial criterion by Rivin describing inscribabilty completely. I think already a cube with corner cut, which is simple, will be a non-inscribable $3$-polytope.

Not all simple polytopes are incribable, e.g. the dual of the cyclic polytope $C_4(8)$ is simple and not inscribable, as shown recently in Combinatorial Inscribability Obstructions for Higher-Dimensional Polytopes by Doolittle, Labbé, Lange, Sinn, Spreer and Ziegler

In dimension $3$, there is a combinatorial criterion by Rivin describing inscribabilty completely. I think already a cube with corner cut, which is simple, will be a non-inscribable $3$-polytope. This can be checked with the following two lines of sage:

sage: C = polytopes.cube().intersection(Polyhedron(ieqs = [[15/8,1,1,1]])) 
....: C.graph().is_inscribable()                                                
False

sage: C.is_simple()                                                             
True

It's nice that Rivin's criterion is implemented in sage...