Let F=(f(k))(k is an element of K) be a Parseval frame in a Hilbert space H, that is, parallel to f parallel to(2) = Sigma(k is an element of K) vertical bar < f,fk >vertical bar(2) holds for every f is an element of H. It is well known that parallel to f(k)parallel to <= 1 for every k, and that if parallel to f(k0)parallel to=1 for some k(0), then f(k0) perpendicular to f(k) for every k not equal K-0. Hence, we might expect that if parallel to f(k0)parallel to is near 1, then the angle between f(k0) and another f(k) is near pi/2, that is, f(k0) and f(k) are not so "close" to each other. We want to make it quantitatively clear by some inequalities. In fact, we can prove the following inequality: vertical bar < f(k), f(l)>vertical bar <= root 1 - parallel to f(k)parallel to(2) center dot root 1 - parallel to f(l)parallel to(2) for k not equal l, which implies that if parallel to f(k)parallel to is near 1 and if parallel to fl parallel to is not so small, then f(k) and f(l) are not so "close" to each other.