# The best constant in an inequality

If $A, B$ are positive definite matrices of order $n$, what is the smallest constant $c$ such that $\det(A^2+B^2)\le c\det(A+B)^2$?

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If such $c$ does not exist, it must depend on the size $n$. – Sunni Jun 1 '12 at 20:16
Have you tried the case where $A$ and $B$ are simultaneously diagonalizable? Might be an easy first step. – Rahul Jun 1 '12 at 20:24
@RahulNarain: When AB=BA, we may take $c=1$. I guess, generally, $c>1$. – Sunni Jun 1 '12 at 20:57

There is no such $c$. Consider the $2 \times 2$ positive definite matrices (for $t \ne 0$) $$A = \pmatrix{1 & 0\cr 0 & t^2\cr}, \ B = \pmatrix{1 & t\cr t & 2 t^2\cr}$$ Then $$\dfrac{\det(A^2 + B^2)}{\det(A+B)^2} = {\dfrac {1+7\,{t}^{2}+{t}^{4}}{25 {t}^{2}}} \to \infty \ \text{as}\ t \to 0$$
Isn't the determinant of $A+B$ in this case $5t^2$, and so the denominator should actually be $25t^4$, giving a constant ratio as $t\rightarrow\infty$? – mjqxxxx Jun 2 '12 at 0:04
Yes $\det(A+B) = 5 t^2$, but $\det(A^2+B^2)$ also has a factor of $t^2$. – Robert Israel Jun 2 '12 at 1:22