On a Norm Inequality for Three 2 × 2 Matrices with One Normal Factor

Abstract

In this paper, we continue to investigate the norm inequality for three real matrices that was recently conjectured by L. László. We establish the validity of the conjecture for the case where $n=2$ and one of the matrices is normal.

1. Introduction

Let $M_n\left(\mathbb{R}\right)$ denote the set of real $n\times n$ matrices. For $A,B\in M_n\left(\mathbb{R}\right)$, Böttcher–Wenzel [1] conjectured that

$${\parallel [A,B]\parallel }_F^2⩽{2\parallel A\parallel }_F^2{\parallel B\parallel }_F^2,$$

(1)

where ${\parallel ·\parallel }_F$ denotes the Frobenius norm. This conjecture has been proved by many authors; for an overview and some generalizations, see [2,3,4,5,6,7] and the references therein. Motivated by (1), László [8] considered three real $n\times n$ matrices and proved a new norm inequality, as follows.

Theorem 1

([8], Theorem 3.1). If $A,B,C\in M_n\left(\mathbb{R}\right)$, then

$${∥ABC-CBA∥}_F^2⩽{∥A∥}_F^2{∥B∥}_F^2{∥C∥}_F^2-{∥B∥}_F^2{\mathrm{tr}}^2\left(A^{T}C\right).$$

(2)

At the same time, László [8] introduced the commutator for $A,B,C\in M_n\left(\mathbb{R}\right)$

$$D:=\left(ABC+BCA+CAB\right)-\left(CBA+ACB+BAC\right),$$

and proposed the following conjecture:

Conjecture 1

([8]). For any $A,B,C\in M_n\left(\mathbb{R}\right)$, the inequality

$${∥D∥}_F^2⩽{\displaystyle \frac{3}{2}}\left({∥A∥}_F^2·{∥[B,C]∥}_F^2+{∥B∥}_F^2·{∥[C,A]∥}_F^2+{∥C∥}_F^2·{∥[A,B]∥}_F^2\right)$$

holds.

László [8] himself verified the conjecture for $n=2$ and $A,B,C$ being upper-triangular. The second cited author [9] established the conjecture for $n=2$, where one of $A,B,C$ is symmetric. In this paper, we will continue to consider László’s conjecture for $n=2$. To be precise, we will prove the following theorem.

Theorem 2.

Let $A,B,C\in M_2\left(\mathbb{R}\right)$, and suppose one of these elements is normal. Consequently,

$${∥D∥}_F^2⩽{\displaystyle \frac{3}{2}}\left({∥A∥}_F^2·{∥[B,C]∥}_F^2+{∥B∥}_F^2·{∥[C,A]∥}_F^2+{∥C∥}_F^2·{∥[A,B]∥}_F^2\right).$$

(3)

Remark 1.

Since every symmetric matrix is normal, our result generalizes Theorem 1.3 in [9].

2. Proof of Theorem 2

Step 1: For $2\times 2$ real matrices, the normal matrices are exactly the symmetric matrices and the scaled rotations of the form $\lambda E_2$ + ${\lambda }_{2}J$, where $E_2$ is the 2 identity matrix and $J=\left(\begin{array}{cc}& 1\\ -1& \end{array}\right)$.

Without loss of generality, we assume C is normal.

(i)

If C is symmetric, then (3) follows from Theorem 1.3 in [9].

(ii)

If C is not symmetric, then

$$C=\left(\begin{array}{cc}{\lambda }_1& {\lambda }_2\\ -{\lambda }_2& {\lambda }_1\end{array}\right)={\lambda }_1{E}_2+{\lambda }_2\left(\begin{array}{cc}& 1\\ -1& \end{array}\right),$$

where ${\lambda }_1,{\lambda }_2\in \mathbb{R}$ and ${\lambda }_2\ne 0$.

Step 2: Computation of $[A,B],[B,C],[C,A]$.

Let

$$A=\left(\begin{array}{cc}{a}_1& a_2\\ a_3& a_4\end{array}\right),B=\left(\begin{array}{cc}{b}_1& b_2\\ b_3& b_4\end{array}\right).$$

Then,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill [A,B]& =\left(\begin{array}{cc}{a}_1& a_2\\ a_3& a_4\end{array}\right)\left(\begin{array}{cc}{b}_1& b_2\\ b_3& b_4\end{array}\right)-\left(\begin{array}{cc}{b}_1& b_2\\ b_3& b_4\end{array}\right)\left(\begin{array}{cc}{a}_1& a_2\\ a_3& a_4\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{a}_1{b}_1+a_2{b}_3& a_1{b}_2+a_2{b}_4\\ a_3{b}_1+a_4{b}_3& a_3{b}_2+a_4{b}_4\end{array}\right)-\left(\begin{array}{cc}{a}_1{b}_1+a_3{b}_2& a_2{b}_1+a_4{b}_2\\ a_1{b}_3+a_3{b}_4& a_2{b}_3+a_4{b}_4\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{a}_2{b}_3-a_3{b}_2& (a_1{b}_2-a_2{b}_1)+(a_2{b}_4-a_4{b}_2)\\ (a_3{b}_1-a_1{b}_3)+(a_4{b}_3-a_3{b}_4)& -(a_2{b}_3-a_3{b}_2)\end{array}\right),\hfill \end{array}\end{array}$$

(4)

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill [B,C]& ={\lambda }_2\left(\begin{array}{cc}-(b_2+b_3)& b_1-b_4\\ b_1-b_4& b_2+b_3\end{array}\right),\hfill \end{array}\end{array}$$

(5)

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill [C,A]& ={\lambda }_2\left(\begin{array}{cc}{a}_2+a_3& a_4-a_1\\ a_4-a_1& -(a_2+a_3)\end{array}\right).\hfill \end{array}\end{array}$$

(6)

Step 3: Computation of $C[A,B],A[B,C],B[C,A]$.

From (4), we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill C[A,B]=& {\lambda }_1\left(\begin{array}{cc}(a_2{b}_3-a_3{b}_2)& (a_1{b}_2-a_2{b}_1)+(a_2{b}_4-a_4{b}_2)\\ -(a_1{b}_3-a_3{b}_1)-(a_3{b}_4-a_4{b}_3)& -(a_2{b}_3-a_3{b}_2)\end{array}\right)\hfill \\ \hfill & -{\lambda }_2\left(\begin{array}{cc}(a_1{b}_3-a_3{b}_1)+(a_3{b}_4-a_4{b}_3)& (a_2{b}_3-a_3{b}_2)\\ (a_2{b}_3-a_3{b}_2)& (a_1{b}_2-a_2{b}_1)+(a_2{b}_4-a_4{b}_2)\end{array}\right).\hfill \end{array}\end{array}$$

(7)

From (5), we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill A[B,C]& =\left(\begin{array}{cc}{a}_1& a_2\\ a_3& a_4\end{array}\right){\lambda }_2\left(\begin{array}{cc}-(b_2+b_3)& b_1-b_4\\ b_1-b_4& b_2+b_3\end{array}\right)\hfill \\ \hfill & ={\lambda }_2\left(\begin{array}{cc}-a_1(b_2+b_3)+a_2\left(b_1-b_4\right)& a_1\left(b_1-b_4\right)+a_2\left(b_2+b_3\right)\\ -a_3(b_2+b_3)+a_4\left(b_1-b_4\right)& a_3\left(b_1-b_4\right)+a_4\left(b_2+b_3\right)\end{array}\right).\hfill \end{array}\end{array}$$

(8)

From (6), we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill B[C,A]& =\left(\begin{array}{cc}{b}_1& b_2\\ b_3& b_4\end{array}\right){\lambda }_2\left(\begin{array}{cc}{a}_2+a_3& a_4-a_1\\ a_4-a_1& -(a_2+a_3)\end{array}\right)\hfill \\ \hfill & ={\lambda }_2\left(\begin{array}{cc}{b}_1(a_2+a_3)+b_2\left(a_4-a_1\right)& b_1\left(a_4-a_1\right)-b_2(a_2+a_3)\\ b_3(a_2+a_3)+b_4\left(a_4-a_1\right)& b_3\left(a_4-a_1\right)-b_4(a_2+a_3)\end{array}\right).\hfill \end{array}\end{array}$$

(9)

Step 4: Computation of ${∥D∥}^2$. Set

$$x_{ij}=a_i{b}_j-a_j{b}_i,1⩽i,j⩽4.$$

It is obvious that $x_{ij}=-x_{ji}$. From (7)–(9), we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill D& =C[A,B]+A[B,C]+B[C,A]\hfill \\ \hfill & ={\lambda }_1\left(\begin{array}{cc}{x}_{23}& x_{12}+x_{24}\\ -x_{13}-x_{34}& -x_{23}\end{array}\right)-{\lambda }_2\left(\begin{array}{cc}2x_{12}+2x_{13}+x_{24}+x_{34}& x_{14}\\ x_{14}& x_{12}+x_{13}+2x_{24}+2x_{34}\end{array}\right),\hfill \end{array}\end{array}$$

(10)

Therefore,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\parallel D\parallel }_F^2& ={\left[x_{23}{\lambda }_1-\left(2x_{12}+2x_{13}+x_{24}+x_{34}\right){\lambda }_2\right]}^2+{\left[\left(x_{12}+x_{24}\right){\lambda }_1-x_{14}{\lambda }_2\right]}^2\hfill \\ \hfill & +{\left[\left(x_{13}+x_{34}\right){\lambda }_1+x_{14}{\lambda }_2\right]}^2+{\left[x_{23}{\lambda }_1+\left(x_{12}+x_{13}+2x_{24}+2x_{34}\right){\lambda }_2\right]}^2\hfill \\ \hfill & =\left[2x_{23}^2+{\left(x_{12}+x_{24}\right)}^2+{\left(x_{13}+x_{34}\right)}^2\right]{\lambda }_1^2\hfill \\ \hfill & +\left[{\left(2x_{12}+2x_{13}+x_{24}+x_{34}\right)}^2+{\left(x_{12}+x_{13}+2x_{24}+2x_{34}\right)}^2+2x_{14}^2\right]{\lambda }_2^2\hfill \\ \hfill & +\left[2x_{23}\left(x_{24}+x_{34}-x_{12}-x_{13}\right)+2x_{14}\left(x_{13}+x_{34}-x_{12}-x_{24}\right)\right]{\lambda }_1{\lambda }_2.\hfill \end{array}\end{array}$$

(11)

Step 5: Computation of ${∥A∥}_F^2·{∥[B,C]∥}_F^2+{∥B∥}_F^2·{∥[C,A]∥}_F^2+{∥C∥}_F^2·{∥[A,B]∥}_F^2$.

From (4)–(6), we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & {∥A∥}_F^2·{∥[B,C]∥}_F^2+{∥B∥}_F^2·{∥[C,A]∥}_F^2+{∥C∥}_F^2·{∥[A,B]∥}_F^2\hfill \\ \hfill =& {2\parallel A\parallel }_F^2\left[{\left(b_2+b_3\right)}^2+{\left(b_1-b_4\right)}^2\right]{\lambda }_2^2+2{\parallel B\parallel }_F^2\left[{\left(a_2+a_3\right)}^2+{\left(a_1-a_4\right)}^2\right]{\lambda }_2^2\hfill \\ \hfill & +2\left({\lambda }_1^2+{\lambda }_2^2\right){\parallel [A,B]\parallel }_F^2.\hfill \end{array}\end{array}$$

(12)

Step 6: Computation of

$$3\left({∥A∥}_F^2·{∥[B,C]∥}_F^2+{∥B∥}_F^2·{∥[C,A]∥}_F^2+{∥C∥}_F^2·{∥[A,B]∥}_F^2\right)-2{∥D∥}_F^2.$$

From (11) and (12), we find that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & 3\left({∥A∥}_F^2·{∥[B,C]∥}_F^2+{∥B∥}_F^2·{∥[C,A]∥}_F^2+{∥C∥}_F^2·{∥[A,B]∥}_F^2\right)-2{∥D∥}_F^2\hfill \\ \hfill =& p{\lambda }_1^2+q{\lambda }_1{\lambda }_2+r{\lambda }_2^2,\hfill \end{array}\end{array}$$

(13)

where

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill p=& {6\parallel [A,B]\parallel }_F^2-2\left[2x_{23}^2+{\left(x_{12}+x_{24}\right)}^2+{\left(x_{13}+x_{34}\right)}^2\right],\hfill \\ \hfill q=& -4\left[x_{23}\left(x_{24}+x_{34}-x_{12}-x_{13}\right)+x_{14}\left(x_{13}+x_{34}-x_{12}-x_{24}\right)\right],\hfill \\ \hfill r=& {6\parallel [A,B]\parallel }_F^2+{6\parallel A\parallel }_F^2\left[{\left(b_2+b_3\right)}^2+{\left(b_1-b_4\right)}^2\right]+6{\parallel B\parallel }_F^2\left[{\left(a_2+a_3\right)}^2+{\left(a_1-a_4\right)}^2\right]\hfill \\ \hfill & -2\left[{\left(2x_{12}+2x_{13}+x_{24}+x_{34}\right)}^2+{\left(x_{12}+x_{13}+2x_{24}+2x_{34}\right)}^2+2x_{14}^2\right]\hfill \end{array}\end{array}$$

(14)

Step 7: Estimation of p. From (4), we have

$${\parallel [A,B]\parallel }_F^2=2x_{23}^2+{\left(x_{12}+x_{24}\right)}^2+{\left(x_{13}+x_{34}\right)}^2.$$

Hence,

$$p=4\left[2x_{23}^2+{\left(x_{12}+x_{24}\right)}^2+{\left(x_{13}+x_{34}\right)}^2\right]⩾0.$$

Step 8: Estimation of

$$3\left({∥A∥}_F^2·{∥[B,C]∥}_F^2+{∥B∥}_F^2·{∥[C,A]∥}_F^2+{∥C∥}_F^2·{∥[A,B]∥}_F^2\right)-2{∥D∥}_F^2$$

under the condition $p=0$.

If $p=0$, then

$$\left\{\begin{array}{c}{x}_{23}=0,\hfill \\ x_{12}+x_{24}=0,\hfill \\ x_{13}+x_{34}=0,\hfill \end{array}\right.$$

which, together with (14), implies

$$\begin{array}{cc}\hfill q=& -4\left[x_{23}\left(x_{24}+x_{34}-x_{12}-x_{13}\right)+x_{14}\left(x_{13}+x_{34}-x_{12}-x_{24}\right)\right]\hfill \\ \hfill =& -4\left[0·\left(x_{24}+x_{34}-x_{12}-x_{13}\right)+x_{14}\left(0-0\right)\right]\hfill \\ \hfill =& 0,\hfill \end{array}$$

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill r=& {6\parallel A\parallel }_F^2\left[{\left(b_2+b_3\right)}^2+{\left(b_1-b_4\right)}^2\right]+6{\parallel B\parallel }_F^2\left[{\left(a_2+a_3\right)}^2+{\left(a_1-a_4\right)}^2\right]\hfill \\ \hfill & -2\left[{\left(x_{12}+x_{13}\right)}^2+{\left(x_{24}+x_{34}\right)}^2+2x_{14}^2\right].\hfill \end{array}\end{array}$$

(15)

Via the AM-GM inequality, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & a_1^2{\left(b_2+b_3\right)}^2+b_1^2{\left(a_2+a_3\right)}^2\hfill \\ \hfill ⩾& {\displaystyle \frac{1}{2}}{\left[a_1(b_2+b_3)-b_1(a_2+a_3)\right]}^2\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{\left[(a_1{b}_2-a_2{b}_1)+(a_1{b}_3-a_3{b}_1)\right]}^2\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{\left(x_{12}+x_{13}\right)}^2,\hfill \end{array}\end{array}$$

(16)

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & a_4^2{\left(b_2+b_3\right)}^2+b_4^2{\left(a_2+a_3\right)}^2\hfill \\ \hfill ⩾& {\displaystyle \frac{1}{2}}{\left[a_4(b_2+b_3)-b_4(a_2+a_3)\right]}^2\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{\left[(a_2{b}_4-a_4{b}_2)+(a_3{b}_4-a_4{b}_3)\right]}^2\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{\left(x_{24}+x_{34}\right)}^2,\hfill \end{array}\end{array}$$

(17)

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & a_1^2{\left(b_1-b_4\right)}^2+b_1^2{\left(a_1-a_4\right)}^2\hfill \\ \hfill ⩾& {\displaystyle \frac{1}{2}}{\left[a_1(b_1-b_4)-b_1(a_1-a_4)\right]}^2\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{\left(a_1{b}_4-a_4{b}_1\right)}^2\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{x}_{14}^2,\hfill \end{array}\end{array}$$

(18)

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & a_4^2{\left(b_1-b_4\right)}^2+b_4^2{\left(a_1-a_4\right)}^2\hfill \\ \hfill ⩾& {\displaystyle \frac{1}{2}}{\left[a_4(b_1-b_4)-b_4(a_1-a_4)\right]}^2\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{\left(a_1{b}_4-a_4{b}_1\right)}^2\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{x}_{14}^2.\hfill \end{array}\end{array}$$

(19)

By combining (15)–(19), we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill r⩾& 6\left[{\displaystyle \frac{1}{2}}{\left(x_{12}+x_{13}\right)}^2+{\displaystyle \frac{1}{2}}{\left(x_{24}+x_{34}\right)}^2+{\displaystyle \frac{1}{2}}{x}_{14}^2++{\displaystyle \frac{1}{2}}{x}_{14}^2\right]\hfill \\ \hfill & -2\left[{\left(x_{12}+x_{13}\right)}^2+{\left(x_{24}+x_{34}\right)}^2+2x_{14}^2\right]\hfill \\ \hfill =& {\left(x_{12}+x_{13}\right)}^2+{\left(x_{24}+x_{34}\right)}^2+2x_{14}^2\hfill \\ \hfill ⩾& 0.\hfill \end{array}\end{array}$$

(20)

Therefore, under the condition $p=0$,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & 3\left({∥A∥}_F^2·{∥[B,C]∥}_F^2+{∥B∥}_F^2·{∥[C,A]∥}_F^2+{∥C∥}_F^2·{∥[A,B]∥}_F^2\right)-2{∥D∥}_F^2\hfill \\ \hfill =& p{\lambda }_1^2+q{\lambda }_1{\lambda }_2+r{\lambda }_2^2\hfill \\ \hfill =& r{\lambda }_2^2\hfill \\ \hfill ⩾& 0.\hfill \end{array}\end{array}$$

(21)

Step 9: Estimation of

$$3\left({∥A∥}_F^2·{∥[B,C]∥}_F^2+{∥B∥}_F^2·{∥[C,A]∥}_F^2+{∥C∥}_F^2·{∥[A,B]∥}_F^2\right)-2{∥D∥}_F^2$$

under the condition $p>0$.

First, we estimate r. Through the Cauchy–Schwartz inequality and AM-GM inequality, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \left[{\left(a_1-a_4\right)}^2+{\left(a_2+a_3\right)}^2\right]\left[{\left(b_2+b_3\right)}^2+{\left(b_1-b_4\right)}^2\right]\hfill \\ \hfill ⩾& {\left[\left(a_1-a_4\right)(b_2+b_3)-\left(b_1-b_4\right)(a_2+a_3)\right]}^2\hfill \\ \hfill =& {\left[(a_1{b}_2-a_2{b}_1)+(a_1{b}_3-a_3{b}_1)+(a_2{b}_4-a_4{b}_2)+(a_3{b}_4-a_4{b}_3)\right]}^2\hfill \\ \hfill =& {\left(x_{12}+x_{13}+x_{24}+x_{34}\right)}^2,\hfill \end{array}\end{array}$$

(22)

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & {\left(a_1+a_4\right)}^2{\left(b_2+b_3\right)}^2+{\left(b_1+b_4\right)}^2{\left(a_2+a_3\right)}^2\hfill \\ \hfill ⩾& {\displaystyle \frac{1}{2}}{\left[\left(a_1+a_4\right)\left(b_2+b_3\right)-\left(b_1+b_4\right)\left(a_2+a_3\right)\right]}^2\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{\left[(a_1{b}_2-a_2{b}_1)+(a_1{b}_3-a_3{b}_1)-(a_2{b}_4-a_4{b}_2)-(a_3{b}_4-a_4{b}_3)\right]}^2\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{\left(x_{12}+x_{13}-x_{24}-x_{34}\right)}^2,\hfill \end{array}\end{array}$$

(23)

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & {\left(a_1+a_4\right)}^2{\left(b_1-b_4\right)}^2+{\left(b_1+b_4\right)}^2{\left(a_1-a_4\right)}^2\hfill \\ \hfill ⩾& {\displaystyle \frac{1}{2}}{\left[\left(a_1+a_4\right)(b_1-b_4)-\left(b_1+b_4\right)(a_1-a_4)\right]}^2\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{\left[2(-a_1{b}_4+a_4{b}_1)\right]}^2\hfill \\ \hfill =& 2x_{14}^2.\hfill \end{array}\end{array}$$

(24)

By combining (22)–(24), we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & {2\parallel A\parallel }_F^2\left[{\left(b_2+b_3\right)}^2+{\left(b_1-b_4\right)}^2\right]+2{\parallel B\parallel }_F^2\left[{\left(a_2+a_3\right)}^2+{\left(a_1-a_4\right)}^2\right]\hfill \\ \hfill =& \left[{\left(a_1+a_4\right)}^2+{\left(a_1-a_4\right)}^2+{\left(a_2+a_3\right)}^2+{\left(a_2-a_3\right)}^2\right]\left[{\left(b_2+b_3\right)}^2+{\left(b_1-b_4\right)}^2\right]\hfill \\ \hfill & +\left[{\left(b_1+b_4\right)}^2+{\left(b_1-b_4\right)}^2+{\left(b_2+b_3\right)}^2+{\left(b_2-b_3\right)}^2\right]\left[{\left(a_2+a_3\right)}^2+{\left(a_1-a_4\right)}^2\right]\hfill \\ \hfill ⩾& 2\left[{\left(a_1-a_4\right)}^2+{\left(a_2+a_3\right)}^2\right]\left[{\left(b_2+b_3\right)}^2+{\left(b_1-b_4\right)}^2\right]\hfill \\ \hfill & +{\left(a_1+a_4\right)}^2{\left(b_2+b_3\right)}^2+{\left(b_1+b_4\right)}^2{\left(a_2+a_3\right)}^2\hfill \\ \hfill & +{\left(a_1+a_4\right)}^2{\left(b_1-b_4\right)}^2+{\left(b_1+b_4\right)}^2{\left(a_1-a_4\right)}^2\hfill \\ \hfill ⩾& 2{\left(x_{12}+x_{13}+x_{24}+x_{34}\right)}^2+{\displaystyle \frac{1}{2}}{\left(x_{12}+x_{13}-x_{24}-x_{34}\right)}^2+2x_{14}^2.\hfill \end{array}\end{array}$$

(25)

Since

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & {\left(2x_{12}+2x_{13}+x_{24}+x_{34}\right)}^2+{\left(x_{12}+x_{13}+2x_{24}+2x_{34}\right)}^2\hfill \\ \hfill =& {\left[\left(x_{12}+x_{13}\right)+\left(x_{12}+x_{13}+x_{24}+x_{34}\right)\right]}^2+{\left[\left(x_{12}+x_{13}+x_{24}+x_{34}\right)+\left(x_{24}+x_{34}\right)\right]}^2\hfill \\ \hfill =& {\left(x_{12}+x_{13}\right)}^2+2\left(x_{12}+x_{13}\right)\left(x_{12}+x_{13}+x_{24}+x_{34}\right)+{\left(x_{12}+x_{13}+x_{24}+x_{34}\right)}^2\hfill \\ \hfill & +{\left(x_{24}+x_{34}\right)}^2+2\left(x_{24}+x_{34}\right)\left(x_{12}+x_{13}+x_{24}+x_{34}\right)+{\left(x_{12}+x_{13}+x_{24}+x_{34}\right)}^2\hfill \\ \hfill =& {\left(x_{12}+x_{13}\right)}^2+{\left(x_{24}+x_{34}\right)}^2+4{\left(x_{12}+x_{13}+x_{24}+x_{34}\right)}^2,\hfill \end{array}\end{array}$$

(26)

together with (26), we find that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill r=& {6\parallel AB-BA\parallel }_F^2+{6\parallel A\parallel }_F^2\left[{\left(b_2+b_3\right)}^2+{\left(b_1-b_4\right)}^2\right]+6{\parallel B\parallel }_F^2\left[{\left(a_2+a_3\right)}^2+{\left(a_1-a_4\right)}^2\right]\hfill \\ \hfill & -2\left[{\left(2x_{12}+2x_{13}+x_{24}+x_{34}\right)}^2+{\left(x_{12}+x_{13}+2x_{24}+2x_{34}\right)}^2+2x_{14}^2\right]\hfill \\ \hfill ⩾& 6{\left(x_{12}+x_{13}\right)}^2+6{\left(x_{24}+x_{34}\right)}^2\hfill \\ \hfill & +6{\left(x_{12}+x_{13}+x_{24}+x_{34}\right)}^2+{\displaystyle \frac{3}{2}}{\left(x_{12}+x_{13}-x_{24}-x_{34}\right)}^2+6x_{14}^2\hfill \\ \hfill & -2\left[{\left(x_{12}+x_{13}\right)}^2+{\left(x_{24}+x_{34}\right)}^2+4{\left(x_{12}+x_{13}+x_{24}+x_{34}\right)}^2+2x_{14}^2\right]\hfill \\ \hfill ⩾& 4{\left(x_{12}+x_{13}\right)}^2+4{\left(x_{24}+x_{34}\right)}^2-2{\left(x_{12}+x_{13}+x_{24}+x_{34}\right)}^2\hfill \\ \hfill & +{\displaystyle \frac{3}{2}}{\left(x_{12}+x_{13}-x_{24}-x_{34}\right)}^2+2x_{14}^2\hfill \\ \hfill ⩾& {\displaystyle \frac{1}{2}}{\left(x_{12}+x_{13}-x_{24}-x_{34}\right)}^2+2x_{14}^2.\hfill \end{array}\end{array}$$

(27)

Secondly, we compute $\mathsf{\Delta }=q^2-4pr$.

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\displaystyle \frac{1}{16}}\mathsf{\Delta }=& {\displaystyle \frac{1}{16}}\left(q^2-4pr\right)\hfill \\ \hfill ⩽& {\left[x_{23}\left(x_{24}+x_{34}-x_{12}-x_{13}\right)+x_{14}\left(x_{13}+x_{34}-x_{12}-x_{24}\right)\right]}^2\hfill \\ \hfill & -\left[2x_{23}^2+{\left(x_{12}+x_{24}\right)}^2+{\left(x_{13}+x_{34}\right)}^2\right]·\left[{\displaystyle \frac{1}{2}}{\left(x_{12}+x_{13}-x_{24}-x_{34}\right)}^2+2x_{14}^2\right]\hfill \\ \hfill ⩽& {\left[x_{23}\left(x_{24}+x_{34}-x_{12}-x_{13}\right)+x_{14}\left(x_{13}+x_{34}-x_{12}-x_{24}\right)\right]}^2\hfill \\ \hfill & -\left[2x_{23}^2+{\displaystyle \frac{1}{2}}{\left(x_{12}+x_{24}-x_{13}-x_{34}\right)}^2\right]·\left[{\displaystyle \frac{1}{2}}{\left(x_{12}+x_{13}-x_{24}-x_{34}\right)}^2+2x_{14}^2\right]\hfill \\ \hfill ⩽& {\left[x_{23}\left(x_{24}+x_{34}-x_{12}-x_{13}\right)+x_{14}\left(x_{13}+x_{34}-x_{12}-x_{24}\right)\right]}^2\hfill \\ \hfill & -{\left[\sqrt{2}{x}_{23}·\sqrt{{\displaystyle \frac{1}{2}}}\left(x_{12}+x_{13}-x_{24}-x_{34}\right)+\sqrt{2}{x}_{14}·{\displaystyle \frac{1}{2}}\left(x_{12}+x_{24}-x_{13}-x_{34}\right)\right]}^2\hfill \\ \hfill =& {\left[x_{23}\left(x_{24}+x_{34}-x_{12}-x_{13}\right)+x_{14}\left(x_{13}+x_{34}-x_{12}-x_{24}\right)\right]}^2\hfill \\ \hfill & -{\left[-x_{23}\left(x_{24}+x_{34}-x_{12}-x_{13}\right)-x_{14}\left(x_{13}+x_{34}-x_{12}-x_{24}\right)\right]}^2\hfill \\ \hfill =& 0.\hfill \end{array}\end{array}$$

(28)

where the first inequality follows from (14) and (27), the second inequality follows from the AM-GM inequality, and the last inequality follows from the Cauchy–Schwartz inequality. Hence,

$$p{\lambda }_1^2+q{\lambda }_1{\lambda }_2+r{\lambda }_2^2⩾0.$$

Based on (13), this means that

$$3\left({∥A∥}_F^2·{∥[B,C]∥}_F^2+{∥B∥}_F^2·{∥[C,A]∥}_F^2+{∥C∥}_F^2·{∥[A,B]∥}_F^2\right)-2{∥D∥}_F^2⩾0.$$

By combining Step 8 and Step 9, we arrive at the following result. Let $A,B,C\in M_2\left(\mathbb{R}\right)$, and suppose one of them is normal. Then,

$${∥D∥}_F^2⩽{\displaystyle \frac{3}{2}}\left({∥A∥}_F^2·{∥[B,C]∥}_F^2+{∥B∥}_F^2·{∥[C,A]∥}_F^2+{∥C∥}_F^2·{∥[A,B]∥}_F^2\right).$$

(29)

Remark 2.

In [9], F. Wang gives the follow example: Let

$$A=\left(\begin{array}{cc}& 1\\ 0& \end{array}\right),B=\left(\begin{array}{cc}& 0\\ 1& \end{array}\right),C=\left(\begin{array}{cc}1& \\ & -1\end{array}\right),$$

Consequently, it is easy to check that

$${∥D∥}_F^2={\displaystyle \frac{3}{2}}\left({∥A∥}_F^2·{∥BC-CB∥}_F^2+{∥B∥}_F^2·{∥CA-AC∥}_F^2+{∥C∥}_F^2·{∥AB-BA∥}_F^2\right)=18,$$

which shows that the constant ${\displaystyle \frac{3}{2}}$ in (29) is optimal.