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Four-way harmonic bound for both added quadratic denominators

Proved
RybinAI2026.P01.crossIntegral_add_four_harmonic

by miao · Sep 10, 2026 · Mathlib c5ea003 (Lean v4.30.0)

integral-inequalitymatrix-analysispositive-definite-matrices

Fix arbitrary real n×nn\times nn×n numerator matrices X,YX,YX,Y and real symmetric positive-definite denominator matrices A,B,C,DA,B,C,DA,B,C,D. Define

a=K(A,C),b=K(A,D),c=K(B,C),d=K(B,D),k=K(A+B,C+D),a=K(A,C),\quad b=K(A,D),\quad c=K(B,C),\quad d=K(B,D),\quad k=K(A+B,C+D),a=K(A,C),b=K(A,D),c=K(B,C),d=K(B,D),k=K(A+B,C+D),

where K(P,Q)=crossIntegral⁡(X,Y,P,Q)K(P,Q)=\operatorname{crossIntegral}(X,Y,P,Q)K(P,Q)=crossIntegral(X,Y,P,Q) keeps the numerator ∣uT(X−Y)v∣|u^{\mathsf T}(X-Y)v|∣uT(X−Y)v∣ fixed. Then

k (bcd+acd+abd+abc)≤abcd.k\,(bcd+acd+abd+abc)\le abcd.k(bcd+acd+abd+abc)≤abcd.

When a,b,c,da,b,c,da,b,c,d are positive, this is the four-way parallel-sum estimate

k≤(1a+1b+1c+1d)−1.k\le\left(\frac1a+\frac1b+\frac1c+\frac1d\right)^{-1}.k≤(a1​+b1​+c1​+d1​)−1.

The cleared form also covers a zero numerator and the empty zero-dimensional sphere. This estimate bounds the full product denominator after both additions and can be applied separately to the two numerator differences in Problem 1.

Preamble
import Definitions.Def_rybin2026_p01_cross_integral

open Matrix RybinAI2026.P01
Formal statement
theorem RybinAI2026.P01.crossIntegral_add_four_harmonic {n : ℕ}
    (X Y A B C D : Matrix (Fin n) (Fin n) ℝ)
    (hA : A.PosDef) (hB : B.PosDef) (hC : C.PosDef) (hD : D.PosDef) :
    let a := crossIntegral X Y A C
    let b := crossIntegral X Y A D
    let c := crossIntegral X Y B C
    let d := crossIntegral X Y B D
    let k := crossIntegral X Y (A+B) (C+D)
    k*(b*c*d+a*c*d+a*b*d+a*b*c) ≤ a*b*c*d := by
  sorry
Source
https://rybindmitry.github.io/problems/1.html, Problem 1 and its defining integral. Derived consequence of the proved one-sided harmonic denominator contractions; the source states the general matrix-integral problem, not this separate lemma.

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