EXACT CLAIM Let A be a Hermitian operator on three qubits with Tr(A)=1 and ≥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|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. 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.