Sorry the title is a bit vague. Let A be a C\*-algebra, and let x and y be positive elements in A. Is it true that $$ \|x-y\|^2 \leq \|x^2-y^2\|? $$
Well, yes. But the proof I have is a bit of a hack, so I wonder if anyone has a "nice" proof, or a reference?


Aside: if $A=C\_0(X)$ then this reduces to the inequality $(a-b)^2 \leq |a^2-b^2|$ for non-negative real numbers a and b.


**Update:** Jonas points me to <http://www.springerlink.com/content/j4756m418220644r/> where Kittaneh has a proof pretty similar to what I had in mind (unpack the proof of Theorem 1). I guess I was interested in whether this sort of thing was standard (if I looked in the right textbook) or if it was a bit of a curiosity. I think the latter seems more likely...



Simple inequality in C*-algebras
Theorem 1.5 of [this 1987 paper](https://dspace.library.uvic.ca:8443/dspace/handle/1828/1506) by J. Phillips says that if $f:[0,\infty)\to [0,\infty)$ is a continuous operator monotone function and $a$ and $b$ are positive operators on a Hilbert space, then $\|f(a)-f(b)\|\leq f(\|a-b\|)-f(0)$. I think that the proof is nice. Corollary 1.6 says that $\|a^{1/n}-b^{1/n}\|\leq\|a-b\|^{1/n}$, $n\geq1.$ Of course your inequality follows from taking $a=x^2$, $b=y^2$, and $n=2$.


Apparently Kittaneh and Kosaki have a similar approach in "Inequalities for the Schatten p-norm. V." Publ. Res. Inst. Math. Sci. 23 (1987), no. 2, 433--443 ([MR link](http://www.ams.org/mathscinet-getitem?mr=890926)). I haven't read any of this article. 


Perhaps I should add the following for a more general audience. A continuous function $f:[0,\infty)\to [0,\infty)$ is operator monotone if whenever $x$ and $y$ are positive operators such that $y-x$ is positive, it follows that $f(y)-f(x)$ is positive. The functions $t\mapsto t^\alpha$ are operator monotone for $0<\alpha\leq1$ (but not for $\alpha>1$).


