From (1le) to stat #
This is (1le) + (beta-bound) ⟹ stat. The right-hand side of (1le),
β(1)/(2α(1-β(1))) + β(1)/(4α) + 1/(2(n-1)) + β(1)/(4α(n-1))
+ a⁴ n(n-1)(n-2) β(1)/(4α) ∫₀¹ t³ β(t)^((n-4)/2) dt,
is monotone increasing in β(1), so (beta-bound) lets β(1) be replaced throughout by
α/(3+α). Under that substitution every term becomes the corresponding term of Sendov.R:
β(1)/(2α(1-β(1))) becomes 1/6, β(1)/(4α) becomes 1/(4(3+α)), and ax, which is
1 - β(1)/2 - α/(n-1), becomes exactly Sendov.c n α.
Monotonicity of the integral term is the only part that is not immediate. ax decreases as
β(1) grows, and Sendov.QQ decreases in its first argument on [0,1], so the integrand
grows — Sendov.integral_QQ_anti. Nonnegativity of β(t) is needed only at the actual value
ax, where it is (ax)² ≤ a², i.e. x² ≤ 1; at the substituted value it comes free from the
pointwise comparison, the same way Sendov.integral_anti gets it in the batching argument.
Main statements #
Sendov.mul_eq_c:ax = 1 - β(1)/2 - α/(n-1), and hencec n α ≤ axunder(beta-bound);Sendov.stat_of_one_le:(1le) + (beta-bound) ⟹ 1 ≤ R n α.
(beta-bound) supplies the feasibility constraint c² ≤ A that Sendov.stat_lt_one
requires. It is c ≤ ax squared, together with x² ≤ 1: the substitution can only lower
ax, and ax is already at most a.
(1le) + (beta-bound) ⟹ stat.