a_507_bc wrote: Do there exist distinct reals $x, y, z$, such that $\frac{1}{x^2+x+1}+\frac{1}{y^2+y+1}+\frac{1}{z^2+z+1}=4$? No, since $\frac 1{x^2+x+1}\le\frac 43$ $\forall x$ with equality when $x=-\frac 12$ and so above statement implies $x=y=z=-\frac 12$

Divide by 4 on both sides and note that eg 4X^2 + 4X +4 = (2X + 1)^2 + 3 Let p= 2X+1, q= 2Y+1, r= 2z+1 and it's to see that the only solution is when p=q=r=0 hence x=y=z=-1/2