Lagrange points and the radius of the Hill sphere
Since the question is about inequalities and ratios, I took the script from this answer and made it better by normalizing it, now $a=1, M1=1$ and $m \equiv M2/M1$

For physical situations in the spirit of Hill spheres and Lagrange points, I think the answer is always going to be:

$$r_1 &lt; r_H &lt; r_2$$

but I can't prove it, that would need math and it's too late tonight for me for math.



Without normalizing $a=1$, the distances and Hill sphere radii for the Sun-Earth and Sun-Mars systems are as follows:

a_Earth:     149598023  km
Sun-Earth L1:  1491524  km
Sun-Earth L2:  1501504  km
Earth r_Hill:  1496531  km

a_Mars:      227939200  km
Sun-Mars L1:   1082311  km
Sun-Mars L2:   1085748  km
Mars r_Hill:   1084032  km


We now apply the normalizations mentioned above:



Python 3 or 2

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import brentq

#unitless   a = 1, M1 = 1, m = M2/M1
# so for the Hill sphere and L1/L2 points of the Earth orbiting the Sun
# m = 5.9724E+24/1.9886E+30  = 4.88987E-07 and
# L1 = 0.0054525 AU, r_Hill = 0.0054625 AU, L2 = 0.0054724 AU

def solve_L1 (r, m):
    return m/r**2 + 1. - r*(1. + m) - (1.-r)**-2

def solve_L2 (r, m):
    return 1 + r*(1 + m) - (1.+r)**-2 - m/r**2

def r_Hill(m):
    return (m / 3.)**(1./3.)

ms = np.logspace(-8, -1, 61)

answers = []
for m in ms:
    r = r_Hill(m)
    L1 = brentq(solve_L1, 0.5*r, 1.5*r, args=(m,))
    L2 = brentq(solve_L2, 0.5*r, 1.5*r, args=(m,))
    answers.append([L2, r, L1])

L2s, rhills, L1s = np.array(list(zip(*answers)))

m_Jup = 1.8982E+27/1.9886E+30
rhill_Jup = r_Hill(m_Jup)
L1_Jup = brentq(solve_L1, 0.5*rhill_Jup, 1.5*rhill_Jup, args=(m_Jup,))
L2_Jup = brentq(solve_L2, 0.5*rhill_Jup, 1.5*rhill_Jup, args=(m_Jup,))

if True:
    plt.figure()
    fs = 14
    plt.subplot(2, 1, 1)
    plt.plot(ms, rhills)
    plt.plot([m_Jup], [rhill_Jup], 'ok')
    plt.text(0.001, 0.03, 'Jupiter', fontsize=fs)
    plt.ylabel('r_Hill / a', fontsize=fs)
    plt.xscale('log')
    plt.yscale('log')
    plt.subplot(2, 1, 2)
    plt.plot(ms, L1s/rhills, '-')
    plt.plot(ms, L2s/rhills, '--')
    plt.plot([m_Jup], [L1_Jup/rhill_Jup], 'ok')
    plt.plot([m_Jup], [L2_Jup/rhill_Jup], 'ok')
    plt.xlabel('m = M2/M1', fontsize=fs)
    plt.ylabel('L1 or L2 / r_Hill', fontsize=fs)
    plt.xscale('log')
    plt.show()