Generalization of a formula for 2x2-matrices It is well known that $$|det(v_1,...,v_n)|\le ||v_1||_2...||v_n||_2$$ 
 with equality if and only if the vectors are pairwise orthogonal. For n = 2,
 the following formula holds : $$det(\pmatrix {a&b\\c&d})^2=(a^2+b^2)(c^2+d^2)-(ac+bd)^2$$
 How can this be generalized to matrices with higher size ?
I checked if $$(a^2+b^2+c^2)(d^2+e^2+f^2)(g^2+h^2+i^2)-det(\pmatrix {a&b&c\\d&e&f\\g&h&i})^2$$ has the factor $$(ad+be+cf)^2+(ag+bh+ci)^2+(dg+eh+fi)^2$$ because the 
 above expression is $0$ if and only if all the scalar products are $0$. But 
 according to wolfram this is not the case, but perhaps there are too many
 variables to check it with the online-version of wolfram.
 A: Let me explain this formula in geometric language. Assume $a=(a_1,a_2),b=(b_1,b_2)$ be two vectors, then the determinant of $\det (a,b)$ is the area of the parallelogram $S$ with two sides $a,b$. Note that
$$S^2:=\left(\det (a,b)\right)^2=(ab_\perp)^2=a^2\left(b^2-\frac{(a \cdot b)^2}{a^2}\right)=a^2b^2-(a\cdot b)^2$$
where $b_\perp$ is the perpendicular component of $b$ with respect to $a$. We uses the Pythagoras theorem in the third identity. 
Similarly in 3 dimension, assume $a,b,c\in\mathbb R^3$, and $S$ the parallelogram (instead of its area) with two sides $a,b$, we have
$$V^2:=(\det(a,b,c))^2=(Sc_\perp)^2=S^2\left(c^2-\frac{(c\cdot S)^2}{S^2}\right)=S^2c^2-(c\cdot S)^2$$
where $c_\perp$ is similarly defined, $c\cdot S=(c\cdot a)b-(c\cdot b)a$ is the projection of $c$ in $S$ up to an area |S|. Replace $S$ by the above, we obtain
$$(\det(a,b,c))^2=a^2b^2c^2-(a\cdot b)^2c^2-(b\cdot c)^2a^2-(c\cdot a)^2b^2+2(a\cdot b)(b\cdot c)(c\cdot a)$$
which is a 3 dimensional generalization. So the essence of the formula is to compute a volume by computing perpendicular components literally. 
A: Notice that if you multiply any of the terms in $ (ad+be+cf)^2+(ag+bh+ci)^2+(dg+eh+fi)^2 $ by a nonzero factor, you get something with the same property. So it might be that there's a factor that looks similar but isn't exactly the same.
Another thing to notice is that you can get the formula $$\det\left(\pmatrix {a&b\\c&d}\right)^2=(a^2+b^2)(c^2+d^2)-(ac+bd)^2$$ in a 'nice' way by writing $$\det\left(\pmatrix {a&b\\c&d}\right)^2=\det\left(\pmatrix {a&b\\c&d}\pmatrix {a&b\\c&d}^T\right)=\det\left(\pmatrix {a^2+b^2&ac+bd\\ac+bd&c^2+d^2}\right)$$
With the $3\times3$ matrix you should get $$\det\left(\pmatrix {a^2+b^2+c^2&ad+be+cf&ag+bh+ci\\ad+be+cf&d^2+e^2+f^2&dg+eh+fi\\ag+bh+ci&dg+eh+fi&g^2+h^2+i^2}\right)$$
Try doing a Laplace expansion on that and see what you get.
