you do lagrange multipliers, f(x)=a+b+c+d=1, finding extrema of g(x)=6*(a^3+...)-(a^2+...)-1/8 grad f=(1,1,1,1) and grad g=(3a^2-2a,3b^2-2b,..) so take l as lambda you get 16 solutions, 2^4 for a +/- for each 3a^2-2a-l=0, 3b^2-2b-l=0, ... you can eliminate a bunch of them, for the rest, solve as a function of l a+b+c+d=1, and (-b)/(2a) for ax^2+bx+c=0 being 1/3, you must have at least one more - than + in +/- sqrt(b^2-4ac), so you have either 4 -'s or 3 -'s, 1+ each of those two cases let you solve for l, and hence for a,b,c,d. so then you just check. theoretically. i'm lazy. it's either true or it isn't. guess i should mention that since cubic grows faster than square, you don't have to worry about a minimum at infinity in any direction * Whtiger has quit IRC ("leaving") so you can take a bound and make it a closed bounded set of values => compact so the minimum will be realized