Under what conditions on a $C^\*$ algebra $A$ we have the following inequality:


$$x^\*a^\*ax+a^\*x^\*xa\leq x^\*x+a^\*x^\*ax+x^\*a^\*xa\;\;\; \forall x,a\in A$$


The second identity which I am looking for is the following:


Does the following inequality imply that the algebra is commutative:


$$xx^\*\leq k x^\*x\;\;\forall x\in A$$ for some positive real $k$? What about if we assume that $k$ is a fixed positive elemnt of the algebra rather than a positive scalar?


The second question is motivated by the fact that every Banach algebra which satisfies $|ab|\leq k|ba|,\;\;\forall a,b$ is necessarily a commutative algebra. The proof is based on Liouville theorem. But in our case the conjucation arising from $\*$ operation destroy holomorphicity.



Two inequalities in $C^*$ algebras
The second condition also implies that $A$ is commutative. If $A$ is not commutative then it has an irreducible representation on some Hilbert space $H$ of dimension at least $2$. Find unit vectors $v,w \in H$ with $\langle v, w\rangle = 0$. By Kadison transitivity there exists $x \in A$ with $xv = 0$ and $xw = v$. Then $\langle x^\*xv, v\rangle = \|xv\|^2 = 0$ but $\langle xx^\*v, v\rangle = \|x^\*v\|^2 \neq 0$ because $\langle x^\*v, w\rangle = \langle v, xw\rangle = 1$. So $xx^\* \leq kx^\*x$ is impossible.


(The second part of the question doesn't really make sense because if $k \in A$ then $kx^\*x$ will not be positive in general. You could ask about $xx^\* \leq x^\*kx$, but this would imply $xx^\* \leq \|k\|x^\*x$ and therefore $A$ must be commutative by the scalar case.)


