 Solution

Take the C-star algebra $M_2(\mathbb{C})$ and chooseThen $a-1=\mathrm{diag}(0,1)$ is positive, whileis positive, since its quadratic form at $(x,y)$ is $|x+y{|}^2$. Thus $1\le a\le b$, as required. However,has determinant $-1$ and eigenvalues $5\pm \sqrt{26}$, one of which is negative. Equivalently, its quadratic form at the real vector $(5,-4)$ is $-4$. It is therefore not positive. This proves that squaring is not order preserving in a C-star algebra, even for invertible elements bounded below by the identity.