Simple inequality in C*-algebras 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...
 A: Theorem 1.5 of this 1987 paper 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).  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 $\alpha>1$).
A: Okay, so actually, decoding Phillip's work gives what I think is a very nice proof.
Let x and y be positive.  Let $\epsilon=\|x-y\|$ so as $x-y$ is self-adjoint, it follows that
$x \leq y+\epsilon 1$.  What seems to be a very standard inequality is that then $x^{1/2} \leq (y+\epsilon 1)^{1/2}$.  We then claim that $(y+\epsilon 1)^{1/2} \leq y^{1/2} + \epsilon^{1/2} 1$.  This follows by working in the commutative C*-algebra generated by y and 1, and using that $(s+t)^{1/2} \leq s^{1/2} + t^{1/2}$ for positive real numbers s and t.  So $$x^{1/2} - y^{1/2} \leq \epsilon^{1/2} 1$$and by symmetry, also $y^{1/2}-x^{1/2}\leq\epsilon^{1/2}1$.  Thus $\|x^{1/2}-y^{1/2}\|\leq\epsilon^{1/2}=\|x-y\|^{1/2}$ as required.
