In fact, we can also say precisely which term has to be non-negative: If w.l.o.g. $x=\min(x,y,z)$, then $x^2+y+\frac{1}{4} \ge x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2 \ge 0$.

Prove that for any positive real numbers $x, y, z$ at least one of numbers $x(y-1) + \frac{1}{4}, y(z-1) + \frac{1}{4}, z(x-1) + \frac{1}{4}$ is nonnegative.

Suppose it is not true, then all the quantities are negatives, then summing them up we conclude that $$x^2 + y+1/4 + y^2+z+1/4 + z^2 + x+1/4<0$$it implies that $(x+1/2)^2 + (y+1/2)^2 + (z+1/2)^2<0$ but it is a contradiction.

Observe that the sum of the given numbers, which can be simplified to be $$\sum_{cyc} \left(x + \frac{1}{2} \right)^2, $$is nonnegative. As such, at least one of the given numbers has to be nonnegative, since if all of them were negative, then the sum would be as well.