Let $x$ and $y$ be positive elements in a C*- algebra $A$. If $c\in A$ such that cx=yc, the prove that $c\sqrt{x}=\sqrt{y}c$.
I could have proved this if it was only $cx=xc$, using continuous functional calculus and approximation by polynomial on the spectrum of $x$. But now I have two elements $x$ and $y$. How should I proceed?
Turning Ryszard Szwarc's comment in an answer:
Since $cx=yx$, use induction to show that $cx^n=y^nc$ for all $n\in\mathbb{N}$. By linearity, we have that $c\cdot p(x)=p(y)\cdot c$ for any polynomial $p(t)$ with $p(0)=0$. Since we can approximate $t\mapsto\sqrt{t}$ uniformly over $[0,M]$ ( where $M\ge\max\{\|x\|,\|y\|\}$ ) by a sequence of polynomials $p_n(t)$ s.t. $p_n(0)=0$, we obtain $c\cdot\sqrt{x}=\sqrt{y}\cdot c$.