Why this inequality related to operators holds

Question

From paper (page 2 ) we have \begin{eqnarray} r\left( AB\pm BA\right) &\leq &\left\Vert AB\right\Vert +\min \left\{ \left\Vert A\right\Vert ^{1/2}\left\Vert AB^{2}\right\Vert ^{1/2},\left\Vert A^{2}B\right\Vert ^{1/2}\left\Vert B\right\Vert ^{1/2}\right\} \label{e.0.9} \\ &\leq &\left\Vert AB\right\Vert +\left\{ \begin{array}{ll} \left\Vert A\right\Vert ^{1/2}\left\Vert B\right\Vert ^{1/2}\left\Vert AB\right\Vert ^{1/2}, & \\ & \\ \min \left\{ \left\Vert A\right\Vert \left\Vert B^{2}\right\Vert ^{1/2},\left\Vert A^{2}\right\Vert ^{1/2}\left\Vert B\right\Vert \right\} , & \end{array}% \right. \notag \end{eqnarray}

I don't understand what the author mean by the second inequality and why it holds

Answer 1

The passage you quote is a direct reference to

F. Kittaneh, Spectral radius inequalities for Hilbert space operators, Proc. Amer. Math. Soc. 134 (2006), 385-390

which I presume you looked at to find an explanation for the bizarre syntax in the quoted passage.

The first line in your quotation is Kittaneh's equation 13 (in Cor. 3), after applying monotonicity of the square root to exchange the order of roots and minimization. The next line is a list of two more quantities no less than the right-hand side of the first line, obtained from the first line by properties of min and norm.

The quoted inequalities are of the form \begin{align*} r(U) &\leq V \\ &\leq ||A B|| + \begin{cases} X , \\ Y, \end{cases} \text{.} \end{align*} The intended meaning is \begin{align*} r(U) &\leq V \text{,} \\ V &\leq ||AB|| + X \text{, and} \\ V &\leq ||AB|| + Y \text{.} \end{align*}

• $r(U) \leq V$ is from Kittaneh.
• From $$||A||^{1/2}||AB^2||^{1/2} \leq ||A||^{1/2}||AB||^{1/2}||B||^{1/2} = X$$ and $$||A^2B||^{1/2}||B||^{1/2} \leq ||A||^{1/2}||AB||^{1/2}||B||^{1/2} = X \text{,}$$ we have that both terms in the min in $V$ are bounded by $X$, so $V \leq ||AB|| + X$.
• If instead, we swap which operator we split out of $||AB^2||$ in the first min expression and out of $||A^2B||$ in the second, we get $Y$.