Inner Product Inequality for unitary vectors Let $(H,\langle,\rangle)$ be an Inner Product Real Space and $\vec a,\vec b,\vec c$ are unitary vectors of $H$. Prove that
$$\vert \langle\vec a,\vec a\rangle\langle\vec b,\vec c\rangle-\langle\vec a,\vec b\rangle\langle\vec a,\vec c\rangle\vert ^2 \leq [\langle\vec a,\vec a\rangle^2-\langle\vec a,\vec b\rangle^2][\langle\vec a,\vec a\rangle^2-\langle\vec a,\vec c\rangle^2]$$
Expanding the LHS and the RHS of I have concluded that it is equivalent to prove that
$$\langle\vec a,\vec b\rangle^2+\langle\vec b,\vec c\rangle^2+\langle\vec c,\vec a\rangle^2\leq 1+2\langle\vec a,\vec b\rangle\langle\vec b,\vec c\rangle\langle\vec a,\vec c\rangle$$
At this point I am stuck. Any idea? Could things have been simplified if I used the Cauchy Schwarz inequality? If the Inner Product Space was Complex then the inequality remains valid?
 A: Suppose that $\dim H \geq 3$. There exists an orthonormal set of vectors $(e_1, e_2, e_3)$ such that :
$$\left\{\begin{array}{lcl}
a & = & e_1 \\[2mm]
b & = & x e_1 + y e_2 + z e_3 \\[2mm]
c & = & u e_1 + v e_2 + w e_3 
\end{array}\right.$$
then :
$$\langle a, b \rangle = x, \langle a, c \rangle = u \text{ and } \langle b, c \rangle = x u + y v + z w$$
$a, b, c$ are unitary than :
$$1 = \langle a, a \rangle = \langle b, b \rangle = \langle c, c \rangle = x^2 + y^2 + z^2 = u^2 + v^2 + z^2 = 1$$
We deduce that :
$$\left(\langle a, a \rangle \, \langle b, c \rangle - \langle a, b \rangle \, \langle a, c \rangle\right)^2 = \left(x u + y v + z w - x u\right)^2 = \left(y v + z w\right)^2$$
and :
$$\left(\langle a, a \rangle^2 - \langle a, b \rangle^2\right) \left(\langle a, a \rangle^2 - \langle a, c \rangle^2\right) = \left(1 - x^2\right) \left(1 - u^2\right) = \left(y^2 + z^2\right) \left(v^2 + w^2\right)$$
We know by the Cauchy-Schwarz inequality that :
$$\left(y v + z w\right)^2 \leq \left(y^2 + z^2\right) \left(v^2 + w^2\right)$$
then :
$$\left(\langle a, a \rangle \, \langle b, c \rangle - \langle a, b \rangle \, \langle a, c \rangle\right)^2 = \left(x u + y v + z w - x u\right)^2 \leq \left(\langle a, a \rangle^2 - \langle a, b \rangle^2\right) \left(\langle a, a \rangle^2 - \langle a, c \rangle^2\right)$$
