32*(2n(n-1)^2 - (1+t)^3) = (t + 1)**2*(t**4 - 2*t**3 + 4*t**2 - 38*t - 29)
Q(t) = t**4 - 2*t**3 + 4*t**2 - 38*t - 29
Q(u+5) = u**4 + 18*u**3 + 124*u**2 + 352*u + 256
coefficients: [1, 18, 124, 352, 256]
all coefficients positive => Q(t) > 0 for all t >= 5  (n >= 7)  [OK]
n=5: 2n(n-1)^2 - (1+sqrt(4n-3))^3 = 108 - 20*sqrt(17) = 108 - 20*sqrt(17) ~ 25.5379
n=6: 2n(n-1)^2 - (1+sqrt(4n-3))^3 = 236 - 24*sqrt(21) = 236 - 24*sqrt(21) ~ 126.0182
n=5,6 endpoints positive exactly  [OK]

TERMINAL INEQUALITY FOR GRAFFITI 290 CERTIFIED SYMBOLICALLY.