A $2\times2$ Matrix inequality

Question

$M,N$ are $2\times2$ real matrices, and $MN=NM$. Then, for any three real numbers $x,y,z$, we have

$$4xz\det(xM^2+yMN+zN^2)\geq(4xz-y^2)\big(x\det(M)-z\det(N)\big)^2 $$

some thought:

1). calculate directly, we got $$ \det(A-xB)=x^2\det(B)-\big(\operatorname{Tr}(A)\operatorname{Tr}(B)-\operatorname{Tr}(AB)\big)x+\det(A) $$ (where $A,B$ are $2\times2$ matrices).

2). $ 4xz\cdot \det(xM^2+yMN+zN^2)= 4x^3z\cdot \det(M-mN)\det(M-nN) $

But I don't know how to go ahead. Thanks a lot!

Answer 1

Had the OP revealed the origin of the question, I might be able to devise a nicer solution. For the time being, I can only solve the problem by a brute-force approach. As $M$ and $N$ commute, there are only three possibilities:

1. At least one of the two matrices is nonsingular. When $M$ is nonsingular, the inequality is equivalent to $$4xz \det(xI + yNM^{-1} + z(NM^{-1})^2) \ge (4xz-y^2) (x-z \det(NM^{-1}))^2.$$ So, we may assume that $M=I$. The inequality then becomes $$4xz\det(xI + yN + zN^2) \ge (4xz-y^2) (x-z \det(N))^2.$$ One can verify that the difference between the two sides is equal to $\left[(x+z\det(N))y + 2xz\operatorname{trace}(N)\right]^2$. Hence the inequality holds.
2. $M,N$ are singular and diagonalisable. We may assume that $M=\operatorname{diag}(m,0)$ and $N=\operatorname{diag}(0,n)$ or $\operatorname{diag}(n,0)$ for some $m,n\in\mathbb R$. When $N=\operatorname{diag}(0,n)$, the inequality reduces to $4x^2z^2m^2n^2\ge0$; when $N=\operatorname{diag}(n,0)$, the inequality reduces to $0\ge0$.
3. $M$ and $N$ are nilpotent matrices. We may assume that they are both scalar multiples of $\pmatrix{0&1\\ 0&0}$. The inequality then reduces to $0\ge0$.

I must stress that this answer is rather unsatisfactory, as it does not really explain how the inequality arises. I hope someone can come up with a more revealing answer.