Square root is operator monotone This is a fact I've used a lot, but how would one actually prove this statement?
Paraphrased: given two positive operators $X, Y \geq 0$, how can you show that $X^2 \leq Y^2 \Rightarrow X \leq Y$ (or that $X \leq Y \Rightarrow \sqrt X \leq \sqrt Y$, but I feel like the first version would be easier to work with)?
Note: $X$ and $Y$ don't have to commute, so $X^2 - Y^2$ is not necessarily $(X+Y)(X-Y)$.

Here is a proof which works more generally for $x,y\ge 0$ in a $C^*$-algebra, where the spectral radius satisfies $r(z)\le \|z\|=\sqrt{\|z^*z\|}$ for every element $z$, and $r(t)=\|t\|$ for every normal element $t$. The main difference with @user1551's argument is that we will use the invertibility of $y$. Other than that, the idea is essentially the same.
If $y$ is not invertible, replace $y$ by $y+\epsilon1$ in the following argument, and then let $\epsilon$ tend to $0$ at the very end.
Now assume that $y$ is invertible and note that
$$x^2\le y^2\quad\Rightarrow\quad y^{-1}x^2y^{-1}\le y^{-1}y^2y^{-1}=1\quad\Rightarrow\quad\left\|y^{-1}x^2y^{-1}\right\|\le 1.$$
Since $t=y^{-1/2}xy^{-1/2}$ and $z=xy^{-1}$ are similar, they have the same spectral radius and therefore
$$\left\|t\right\|=r(t)=r\left(z\right)\le\|z\|=\sqrt{\|z^*z\|}=\sqrt{\left\|y^{-1}x^2y^{-1}\right\|}\le 1.$$
It follows that $y^{-1/2}xy^{-1/2}\le 1$, whence $x\le y$ as desired.
