If and all are distinct such that:
,then show that are in G.P.
The discriminant is
which is never positive by the Cauchy–Schwarz inequality. Hence, since the coefficient of is never negative, we have
Combining this with the given condition, we find that the quadratic expression is equal to zero. Hence the discriminant is zero, so we have equality for Cauchy–Schwarz:
This means that the vectors and are linearly dependent, i.e. such that Hence the ’s are in geometric progression.