To prove this, let us consider, as in the December riddle, a weighted average
between the number of appearances of L_{2}, K_{2,3}
and T^{+}, which we'll henceforth call our "usual suspects".
(Here and throughout, we use terminology defined in the
December 2015 riddle and its solution.) This time we will use the following
weighing of these three graphs:

For each *i* value, the number of appearances of C_{4} is
*X _{i}i*/4, so the ratio between the number of C

The ratio over all *i* is a weighted average over these local ratios, so
it, too, cannot be greater than 5/6.

To describe a graph *G* for which this ratio is exactly 5/6, let us
first consider three simpler graphs. The first graph is
"C_{4} x C_{3}".
The notation "*G* x *H*" for two graphs *G* and *H*
indicates the graph whose vertices are the pairs

and whose edges are

The other two are K_{4,4} and K_{5,5}, where
K_{n,n} is the complete bipartite graph with *n*
vertices on each side.

We use these graphs because in each of them all *X _{i}* are zero
except

Our target graph, *G* will be the following:

The structure of this 4080-vertex graph makes the number of appearances in any
subgraph of
it quite easy to compute. In particular, the numbers have been designed so that
all of our usual suspects will have exactly the same number of appearances,
32,400, meaning that the number of appearances of C_{4},
27,000, is exactly at the desired ratio, 5/6, of this number.

To complete the proof, however, we still need to show that our usual suspects
are the most frequent connected induced subgraphs of *G*.
To prove this, note that all
subgraphs of K_{4,4} and K_{5,5} are bipartite, while almost
all connected induced subgraphs of C_{4} x C_{3} except
the single vertex and single edge subgraphs include a triangle, so cannot be
bipartite. The subgraphs that do not appear in both types of components will
certainly not appear more than the maximum frequency connected induced
subgraphs of these components, which are our usual suspects.

All in all, there are 5 subgraphs of *G* that have the same maximal
frequency, 32,400. These are our three usual suspects, a 6-node subgraph with
the adjacency matrix

011110 101110 110001 110011 110101 001110

and a 7-node subgraph with the adjacency matrix

0110011 1011100 1101100 0110111 0111011 1001101 1001110

Q.E.D.

I am not aware of any *connected* *G* that satisfies the criteria
of this riddle. Any reader who discovers such a *G* is most welcome to
send me its description.