# An operator inequality

I would be most thankful if you could help me prove the following operator inequality. Let $A$ be an arbitrary linear operator on a Hilbert space, satisfying $$\left\|AA^{\ast} - A^{\ast}A\right\|\leq 2a$$ where $A^{\ast}$ is the Hermitian adjoint and $a>0$ is a constant. Let $\varepsilon$ be equal to either $+1$ or $-1$. Then show that $$2\sqrt{A^{\ast}A + aI} - \varepsilon\left(A + A^{\ast}\right) \geq 0$$ Thank you!

In this answer we assume that $A$ is a bounded operator on a complex Hilbert space $H$.

Sketched proof and hints:

1. We may uniquely write $A=B+iC$, where $B$ and $C$ are selfadjoint operators. Define selfadjoint operator $D:=i[C,B]=D^{\dagger}$. Then OP's statement reads $$\tag{1} ||D|| ~\leq ~a \qquad \Rightarrow \qquad \sqrt{A^{\dagger}A+aI} ~\geq~ \pm B.$$

2. Show that (1) is a consequence of $$\tag{2}||D|| ~\leq ~a \qquad \Rightarrow \qquad\sqrt{A^{\dagger}A+aI}~\geq~ \sqrt{ B^2 }.$$

3. Show that (2) is a consequence of $$\tag{3}||D|| ~\leq ~a \qquad \Rightarrow \qquad\sqrt{A^{\dagger}A+aI}~\geq~ \sqrt{ B^2 +C^2 }.$$

4. Show that (3) is equivalent to $$\tag{4}||D|| ~\leq ~a \qquad \Rightarrow \qquad A^{\dagger}A+aI~\geq~ B^2 +C^2 .$$

5. Show that (4) is equivalent to $$\tag{5}||D|| ~\leq ~a \qquad \Rightarrow \qquad aI~\geq~ D .$$

6. Show (5).