EXACT CLAIM

Let A be a Hermitian operator on three qubits with Tr(A)=1 and <s|A|s>≥0 for every pure three-qubit stabilizer state. Suppose A=cI+aP+bQ for orthogonal rank-one projectors P,Q and arbitrary real a,b,c. Then Tr(A²)≤2. Equality holds if and only if A is Clifford-conjugate to [(I+X+Y+Z)/2]⊗|00><00|. In particular the bound is sharp on the entire multiplicity-(1,1,6) spectral sector, including all eigenvalue orderings and their coalescences. This is not the unrestricted three-qubit or all-n inradius conjecture.

FULL FROZEN PROOF

Part I. Complete same-sign exclusion (the native round40 initial proof, reproduced with its existing notation).
The proof in this part assumes ab≥0. Its conclusion is the strict bound Tr(A²)<2 in that sector. The definitions and compression identities before the use of ab≥0 will also be used in Part II.

We use the established two-qubit inradius result supplied in the problem: the Hilbert–Schmidt ball of radius 1/√28 about I_4/4 is contained in STAB_2. First, any Hermitian two-qubit stabilizer-positive B, with p=Tr B, satisfies
Tr(B²)≤2p².                                                     (1)
Indeed, summing over a stabilizer basis gives p≥0. Applying stabilizer positivity to the ball point opposite B−pI_4/4 gives ||B−pI_4/4||_2≤p√28/4. Squaring and adding p²/4 proves (1), including p=0. This applies equally to compressions into either eigenspace of a nonidentity three-qubit Pauli: a Clifford identification makes that eigenspace a two-qubit stabilizer code, whose encoded stabilizer states are three-qubit stabilizer states.

It suffices to exclude Tr(A²)=2. In fact, if an admissible A had purity greater than 2, the segment from I/8 to A would contain an admissible operator of purity exactly 2. This interpolation multiplies a,b by the same positive number, preserving ab≥0 and the projectors.

Suppose, therefore, that Tr(A²)=2. Put s=a+b and δ=a−b. The trace and purity equations give
c=(1−s)/8,     3s²+4δ²=15.                                    (2)
Since ab=(s²−δ²)/4≥0, equation (2) implies
15/7≤s²≤5.                                                    (3)

Take the 63 nonidentity standard Hermitian three-qubit Paulis T. Write P=|ψ⟩⟨ψ|, Q=|φ⟩⟨φ|, and set
u=⟨ψ|T|ψ⟩, v=⟨φ|T|φ⟩, z=|⟨ψ|T|φ⟩|²,
t=au+bv, q=au−bv.
For E_±=(I±T)/2, the trace of the compression B_±=E_±AE_± on its four-dimensional range is p_±=(1±t)/2. These compressions are stabilizer-positive, so p_±≥0 and hence |t|≤1. Expanding (1) for both compressions, and using (2), yields
D≥|L|,
D=(s²−1)/32+(3t²−q²)/8−abz/2,
L=(7−s)t/8−δq/4.                                             (4)
For clarity, the expansion uses
Tr(B_±²)=4c²+c(s±t)+[a²(1±u)²+b²(1±v)²+2abz]/4;
thus 2p_±²−Tr(B_±²)=D±L.

Changing the sign of T if necessary makes t≥0 without changing D or |L|. Completing the square gives
D−L=(7−s²)/16+[3t²−(7−s)t]/8−(q−δ)²/8−abz/2.
Since ab≥0 and D−L≥0, necessarily
6t²−2(7−s)t+7−s²≥0.                                         (5)
The quadratic in (5), evaluated at t=1, equals −(s−1)²<0 by (3). Its roots are
[7−s±√7|s−1|]/6.
Since 0≤t≤1, it follows that, for every Pauli,
|t|≤T(s):=[7−s−√7|s−1|]/6.                                  (6)

The orthogonality of ψ and φ and Pauli Parseval give
Σu²=Σv²=7, Σuv=−1, Σz=8.
Consequently
Σt²=3s²+4δ²=15,
Σq²=4s²+3δ²,   Σtq=7sδ,
ΣD=6,
ΣL²=S(s):=[−105s⁴+588s³+450s²−3780s+3615]/256.                (7)
All sums here run over the 63 nonidentity Paulis. In particular, the formula for S follows directly from
ΣL²=15(7−s)²/64+δ²(11s²−49s+3δ²)/16
and δ²=(15−3s²)/4.

First consider s<0, which is precisely the both-nonpositive-offset case under (3). Since s≤−√(15/7)<−1, (6) gives
|t|≤(4−√7)/3<11/24,
where √7>21/8 follows from 7−(21/8)²=7/64>0. Therefore
Σt²<63(11/24)²=847/64<15,
contradicting (7). The last gap is exactly 15−847/64=113/64>0.

It remains to consider s>0. By (3), √(15/7)≤s≤√5, so s>1. Dropping the nonpositive terms in D gives, by (6),
0≤D≤(s²−1)/32+3T(s)²/8.
Use √7>21/8 and s>1 to obtain
0≤T(s)≤(77−29s)/48.
The right side is positive on the interval in question (it is already positive at s=9/4). Thus
D≤M(s):=(1033s²−4466s+5737)/6144.
Combining this with (4) and (7) forces
S(s)=ΣL²≤ΣD²≤M(s)ΣD=6M(s).                                  (8)
We now certify the opposite strict inequality on the entire interval, rather than only at sampled points. Direct subtraction gives
F(s):=S(s)−6M(s)
=[−420s⁴+2352s³+767s²−10654s+8723]/1024.
The interval [√(15/7),√5] is contained in [73/50,9/4], since
15/7−(73/50)²=197/17500>0,   (9/4)²−5=1/16>0.
Write s=73/50+(79/100)x, with 0≤x≤1. The degree-four Bernstein coefficients of F in this coordinate are exactly
33512387/160000000,
122637237/256000000,
81552893/61440000,
1123221/409600,
298307/65536.
Each is strictly positive. The Bernstein basis functions binom(4,i)x^i(1−x)^(4−i) are nonnegative and sum to one, proving F(s)>0 throughout that interval. This contradicts (8).

Both signs allowed by (3) have been excluded. Hence purity 2 is impossible, and the interpolation argument excludes every larger purity as well. This proves the stated strict inequality for all ab≥0.

Part II. The mixed-sign sector.
Suppose ab<0 and first assume Tr(A²)=2. Retain the definitions s,delta,u,v,z,t,q,D,L,S from Part I. Its equations (2),(4),(7) hold for arbitrary real a,b; the sign restriction was only used after those identities. Now |s|<sqrt(15/7)<3/2 and delta²=(15-3s²)/4.

The compression of a Hermitian Pauli T to W=ran(P+Q) is the Hermitian contraction [[u,h],[conj(h),v]] with |h|²=z. Positivity of both I plus and I minus this compression gives
z≤min((1-u)(1-v),(1+u)(1+v))=1+uv-|u+v|.
Put w=s*t-delta*q=2ab(u+v). Because -ab/2>0, substituting this upper bound into D in the correct direction gives
D≤(7-3s²)/16+t²/4-|w|/4,
L=(7-3s)t/8+w/4.
Thus D≥|L| and the triangle inequality imply
4t²-2(7-3s)|t|+7-3s²≥0.                          (9)
Here 7-3s>0. The left side at |t|=1 is -3(s-1)², and |t|≤1 already follows from Pauli-compression trace positivity. Its discriminant is84(s-1)². For s≠1 the number1 lies strictly between its roots, so the allowed values must lie below the smaller root. At s=1 the same bound is simply |t|≤1. Therefore in all cases
|t|≤T(s):=[7-3s-sqrt(21)|s-1|]/4.                 (10)

Use sqrt(21)>9/2, whose squared gap is3/4>0. On -3/2≤s≤1 we obtain T(s)≤(5+3s)/8; on1≤s≤3/2 we obtain T(s)≤(23-15s)/8. These last upper bounds are positive on their stated intervals. Omitting -|w|/4 and using (10), we get
0≤D≤M_-(s):=(137+30s-39s²)/256     for s≤1,
0≤D≤M_+(s):=(641-690s+177s²)/256   for s≥1.
Both M functions are positive on the relevant intervals: the first is concave with positive endpoint values, and the second is decreasing there with value17/1024 at3/2. Consequently the Parseval sums imply
S(s)=sum L²≤sum D²≤6M_±(s).
But exact factorization gives
256[S-6M_-]=(1-s)[105s³-483s²-1167s+2793],
256[S-6M_+]=(s-1)[-105s³+483s²-129s+231].
For -3/2≤s≤0 the first bracket is at least10815/8; for0≤s≤1 it is at least1143. These estimates follow by replacing negative cubic and quadratic terms by their endpoint lower bounds and nonnegative terms by zero. For1≤s≤3/2 the second bracket is at least1329/8, using s³≤27/8,s²≥1,s≤3/2. All three rational bounds are strictly positive. This contradicts S≤6M unless s=1.

At s=1 we have c=0,delta²=3,ab=-1/2, and
0≤D≤1/4+t²/4-|w|/4≤1/2,
sum L²=3, sum D=6.
Therefore equality holds throughout
3=sum L²≤sum D²≤(1/2)sum D=3.
Termwise, each D is either0 or1/2, exactly twelve of them equal1/2, and for each of these twelve Paulis |t|=1 and w=0. This already excludes every admissible A of purity greater than2: radial interpolation B=(1-k)I/8+kA reaches purity2 at some0<k<1 and preserves the signs ofa,b. Part I excludes the same-sign case; Part II forces aPauli with|Tr(TB)|=1. But any trace-one stabilizer-positive A has|Tr(TA)|≤1, since each Pauli eigenspace has a stabilizer basis. As Tr(TB)=kTr(TA), this is impossible.

Part III. Equality classification.
Let Tr(A²)=2. Part I excludes ab≥0 and Part II gives s=1,c=0. Thus A is supported on the two-dimensional space W=ran(P+Q). For each of the twelve Paulis with D=1/2, equality in the displayed upper bound gives |t|=1,w=0. Since ab=-1/2 is nonzero, w=2ab(u+v)=0 implies v=-u. Hence t=delta*u and u²=1/3. Equality in the contraction bound is necessary, because D=1/2 saturates the upper bound and its z coefficient is strictly positive. Thus z=1-u²=2/3. The compressed2×2 matrix of T on W has trace0 and squareI. If R is the orthogonal projector onto W, the identity
R-(RTR)²=RT(I-R)TR
then shows (I-R)TR=0. Every one of these twelve distinct Pauli labels preserves W.

Let H be the subgroup of phase-free Pauli labels preserving W. It is a vector subspace of the six-dimensional binary symplectic Pauli label space: products preserve W, phases are irrelevant, and each Pauli is invertible. It contains these twelve labels and the identity, hence |H|≥13 and dim H≥4. An isotropic subspace of the six-dimensional nondegenerate symplectic space has dimension at most3; thus H contains an anticommuting pair U,V. Restricted to W they are anticommuting Hermitian involutions on dimension2, and generate the full2×2 matrix algebra.

The subgroup H0 of H commuting with both U and V has dimension dimH-2, because their symplectic pairings give two independent linear functionals onH. Every member ofH0 restricts to a scalar ±I onW: it commutes with the irreducible pair. Members ofH0 must commute with each other globally, for anticommuting Pauli operators could not have simultaneous nonzero scalar restrictions toW. Choose independent Hermitian representatives with their signs fixed to act as+I onW. Their joint+1 eigenspace has dimension2^(3-dimH0). Since it contains W of dimension2, dimH0≤2. Combining this with dimH≥4 gives dimH=4 and dimH0=2; W is exactly a two-generator, rank-two stabilizer code.

The standard stabilizer normal form, obtained by symplectic elimination of two independent commuting Pauli labels, supplies a Clifford sending W to C²⊗|00>. Under that Clifford, A becomes A_L⊗|00><00| with Tr(A_L)=1. Write A_L=(I+r_xX+r_yY+r_zZ)/2. Each encoded one-qubit stabilizer state is a three-qubit stabilizer state, so |r_x|,|r_y|,|r_z|≤1. Purity2 gives r_x²+r_y²+r_z²=3, forcing all three components to be±1. The single-qubit Clifford group permutes these eight cube corners transitively, proving the claimed necessity.

Conversely, take A0=[(I+X+Y+Z)/2]⊗|00><00|. Projection of a stabilizer state onto the computational outcome00 on the last two qubits is either zero or an unnormalized one-qubit stabilizer state. This follows by successive Pauli-Z measurements: a commuting stabilizer already fixes the outcome, or an anticommuting generator can be replaced by the measured signed Pauli, preserving a complete commuting generating set; after both measurements the residual pure state is stabilized by a one-qubit Pauli. The expectation of(I+X+Y+Z)/2 on each of the six one-qubit stabilizer states is0or1, hence A0 is stabilizer-positive. Its trace is1 and purity2, with eigenvalues(1±sqrt3)/2 and six zero eigenvalues. Clifford conjugation preserves all these properties, finishing sufficiency.

Provenance and scope. Part I is the complete native initial claim. Part II is the independently derived current-round root lead. The equality route was developed in the native followup STATE and independently completed by the root assistant here. That followup timed out on its final request without submitting a complete draft; its NO_CLAIM and unknown usage are retained. This is a newly assembled complete root candidate requiring its own frozen review, not a retroactive native submission. The code checks exact scalar identities and sharpness identities; the analytic proof supplies the universal geometry and equality classification. No formal proof assistant, literature priority or external expert confirmation is claimed.


REMAINING GAP

Arbitrary other spectra, unrestricted three-qubit stabilizer-positive operators and the all-n inradius conjecture remain open. Literature novelty and external review have not been established.
