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Let $A, B, C$ be $2\times 2$ Hermitian positive semidefinite (PSD) matrices. Prove or disprove that $$\det(6(A^3 + B^3 + C^3)+(A+B+C)^3) \ge 5^2\det(A^2 + B^2 + C^2)\det(A + B+ C).$$

This problem is inspired by these two problems: P1, P2. When I tried to deal with P2, I want to eliminate the constraint $A+B+C = I_2$. So I propose this problem.

I just do some numerical experiments and do not find an counterexample yet.

We can use the fact that for $2\times 2$ matrix $S$, $$\det S = \frac12 (\mathrm{tr}(S))^2 - \frac12\mathrm{tr}(S^2).$$

An idea is to use $A = u_1u_1^H + u_2u_2^H, B = v_1v_1^H + v_2v_2^H, C = w_1w_1^H + w_2w_2^H$. Then we have a unconstrained optimization of $f(u_1, u_2, v_1, v_2, w_1, w_2)$.

I also consider the following:

Let $A, B$ be $2\times 2$ Hermitian positive semidefinite (PSD) matrices. Prove or disprove that $$\det(6(A^3 + B^3 + I_2)+(A+B+I_2)^3) \ge 5^2\det(A^2 + B^2 + I_2)\det(A + B+ I_2).$$

River Li
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1 Answers1

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In general need to prove: $$\det(6(A^3+B^3+I) + (A+B+I)^3) \geq 25 \times \det((A^2+B^2+I)(A+B+I))$$

If $AB = BA$, we can assume $A = D$, $B = D'$, diagonal matrices and the inequality is true iff for all $x \geq 0$, $y \geq 0$, $$6(x^3+y^3+1)+(x+y+1)^3 \geq 5 \times (x^2+y^2+1)(x+y+1)$$ $$6 \left( (x+y+1)^3-3x^2y-3y^2x-6xy-3(x+y) \right)+(x+y+1)^3 \geq 5 (x^2+y^2+1)(x+y+1)$$ $$\left( 7(x+y+1)^3-18xy(x+y)-36xy-18(x+y) \right) \geq 5 (x^2+y^2+1)(x+y+1)$$ $$\left( 7/5 (x+y+1)^2-\frac{18xy(x+y)}{5(x+y+1)}-\frac{36xy}{5(x+y+1)}-\frac{18(x+y)}{5(x+y+1)} \right) \geq (x^2+y^2+1)$$ $$\left( \frac{2((x-y)^2+1)}{5} + \frac{14(x+y)}{5}-\frac{18xy}{5(x+y+1)}-\frac{18}{5}+\frac{18}{5(x+y+1)} \right) \geq 0$$ $$\left( \frac{2((x-y)^2+1)}{5} + \frac{14(x^2+y^2+x+y)}{5(x+y+1)}+\frac{10xy}{5(x+y+1)}-\frac{18}{5}+\frac{18}{5(x+y+1)} \right) \geq 0$$ $$\left( \frac{2(x-y)^2}{5} + \frac{14(x^2+y^2-1)}{5(x+y+1)}+\frac{10xy}{5(x+y+1)}-\frac{2}{5}+\frac{18}{5(x+y+1)} \right) \geq 0$$

which is true for all $x \geq 0, y \geq 0$.

Balaji sb
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