I am currently working on a problem in differential geometry and came across the following calculation which I can't seem to solve. Let $$v_1 = \langle b_1,b_2,b_3,b_4\rangle \;\; and \;\; v_2=\langle c_1,c_2,c_3,c_4 \rangle$$ be real orthonormal vectors in $\mathbb{R}^4$. Prove that $$|c_1b_2-c_2b_1+ c_3b_4-b_3c_4|\leq 1$$ It is not clear to me how to prove this result. All that I have noticed is that I can build a semi-orthogonal matrix $A = (v_2\;v_1)$, where $v_1$ and $v_2$ are column vectors. The above calculation is then seen to pertain to the determinants of the $2\times 2$ submatrices.
Any help on this problem would be greatly appreciated.
You have that they are orthonormal so both of those vectors have unit norm. Then:
$$|c_1b_2-c_2b_1 +c_3 b_4 - b_3 c_4| \leq |c_1 b_2| + |c_2 b_1| + |c_3 b_4| + |b_3 c_4|$$
by the triangle inequality. Now, apply the Cauchy-Schwarz Inequality to the right-hand side and you will get that:
$$|c_1 b_2| + |c_2 b_1| + |c_3 b_4| + |b_3 c_4| \leq \sqrt{c_1^2+c_2^2+c_3^2+c_4^2} \sqrt{b_1^2+b_2^2+b_3^2+b_4^2} = 1 \cdot 1 = 1$$
and we are done. The orthogonality of both vectors does not play a role in the argument.