Theorem: Square Root is Increasing with respect to Canonical Preordering of C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $\le_A$ be the canonical preordering of $A$. 
Let $x, y \in A$ be positive and such that:
:$x \le_A y$
Let $x^{1/2}$ and $y^{1/2}$ be the positive square roots of $a$ and $b$ respectively.
Then:
:$x^{1/2} \le_A y^{1/2}$

Proof:
Write $a = x^{1/2}$ and $y^{1/2}$. 
We then have $a^2 \le_A b^2$ and want to show that $a \le_A b$. 
First take $A$ unital.
Let $t > 0$.
Let:
:$u = \paren {t {\mathbf 1}_A + b + a} \paren {t {\mathbf 1}_A + b - a}$ 
From Element of *-Algebra Uniquely Decomposes into Hermitian Elements, there exists unique Hermitian elements $c, d \in A$ such that:
:$u = c + i d$
with:
:$\ds c = \frac 1 2 \paren {u^\ast + u}$
We have:
:$u^\ast = \paren {t {\mathbf 1}_A + b - a} \paren {t {\mathbf 1}_A + b + a}$
from $(\text C^\ast 3)$ in the definition of an involution and Identity Element in Unital *-Algebra is Hermitian.
Hence we have:
{{begin-eqn}}
{{eqn	| l = c
		| r = \frac 1 2 \paren {\paren {t {\mathbf 1}_A + b + a} \paren {t {\mathbf 1}_A + b - a} + \paren {t {\mathbf 1}_A + b - a} \paren {t {\mathbf 1}_A + b + a} }
}}
{{eqn	| r = \frac 1 2 \paren {t^2 {\mathbf 1}_a + t \paren {b + a} + t \paren {b - a} + \paren {b + a} \paren {b - a} + t^2 {\mathbf 1}_a + t \paren {b - a} + t \paren {b + a} + \paren {b - a} \paren {b + a} }
}}
{{eqn	| r = \frac 1 2 \paren {2 t^2 {\mathbf 1}_A + 4 t b + b^2 + a b - b a - a^2 + b^2 - a b + b a - a^2}
}}
{{eqn	| r = \frac 1 2 \paren {2 t^2 {\mathbf 1}_A + 4 t b + 2 b^2 - 2 a^2}
}}
{{eqn	| r = t^2 {\mathbf 1}_A + 2 t b + b^2 - a^2 
}}
{{end-eqn}}
From Non-Negative Multiple of Positive Element of C*-Algebra is Positive, we have that $2 t b$ is positive.
Since $a^2 \le_A b^2$, we also have that $b^2 - a^2$ is positive.
Hence:
:$t^2 {\mathbf 1}_A + 2 t b + b^2 - a^2 \ge_A t^2 {\mathbf 1}_A$
so that $c \ge_A t^2 {\mathbf 1}_A$.
From Element of Unital C*-Algebra Bounded below by Non-Negative Multiple of Identity Element is Invertible and Positive, we have that $c$ is invertible and positive.
We can therefore define the square root $c^{1/2}$.
Since $c^{1/2}$ commutes with itself and $c = c^{1/2} c^{1/2}$ with $c$ invertible, we have that $c^{1/2}$ is invertible by Product of Commuting Elements in Monoid is Unit iff Each Element is Unit.
Write $c^{-1/2} = \paren {c^{1/2} }^{-1}$.
We have that $c^{-1/2} d c^{-1/2}$ is Hermitian by Star of Product of Elements in *-Algebra.
Hence from Spectrum of Hermitian Element in Unital C*-Algebra is Real, we have:
:$\map {\sigma_A} {c^{-1/2} d c^{-1/2} } \subseteq \R$
From Spectral Mapping Theorem for Polynomials, we have:
:$\map {\sigma_A} {i c^{-1/2} d c^{-1/2} } \subseteq i \R = \set {t i : t \in \R}$
In particular, $-1 \not \in \map {\sigma_A} {i c^{-1/2} d c^{-1/2} }$.
Hence ${\mathbf 1}_A + i c^{-1/2} d c^{-1/2}$ is invertible.
We can then write:
:${\mathbf 1}_A + i c^{-1/2} d c^{-1/2} = c^{-1/2} \paren {c + i d} c^{-1/2} = c^{-1/2} u c^{-1/2}$
Hence from Inverse of Product, we have that:
:$u$ is invertible
since $\paren {c^{-1/2} }^{-1} = c^{1/2}$.
So $\paren {t + b + a} \paren {t + b - a}$ is invertible.
From Sum of Two Positive Elements of C*-Algebra is Positive, we have that $b + a$ is positive.
Hence:
:$\map {\sigma_A} {b + a} \subseteq \hointr 0 \infty$ 
In particular:
:$-t \not \in \map {\sigma_A} {b + a}$ 
Hence $t + b + a$ is invertible.
Hence by Inverse of Product:
:$t + b - a = \paren {t + b + a}^{-1} \paren {\paren {t + b + a} \paren {t + b - a} }$ is invertible.
So $-t \not \in \map {\sigma_A} {b - a}$.
Since $t > 0$ was arbitary we have:
:$\map {\sigma_A} {b - a} \cap \openint {-\infty} 0 = \O$
Since $b - a$ is Hermitian, we have:
:$\map {\sigma_A} {b - a} \subseteq \R$ 
from Spectrum of Hermitian Element in Unital C*-Algebra is Real.
Hence we have:
:$\map {\sigma_A} {b - a} \subseteq \hointr 0 \infty$
So $a \le_A b$.
Now suppose that $A$ is not unital.
Let $A_+$ be the unitization of $A$. 
Let $\le_+$ be the canonical preordering on $A_+$.
Define:
:$\tuple {u, \lambda}^\ast = \tuple {u^\ast, \overline \lambda}$
for each $\tuple {u, \lambda} \in A_+$.
From Existence of Unitization of C*-Algebra, there exists a unique algebra norm $\norm {\, \cdot \,}_\ast$ extending $\norm {\, \cdot \,}$ such that $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ is a unital $\text C^\ast$-algebra.
From Element of C*-Algebra is Positive iff Positive in Unitization, we have:
:$\tuple {x, 0} \le_+ \tuple {y, 0}$
Hence by the unital case, we have:
:$\tuple {x, 0}^{1/2} \le_+ \tuple {y, 0}$
That is:
:$\tuple {x^{1/2}, 0} \le_+ \tuple {y^{1/2}, 0}$
From Element of C*-Algebra is Positive iff Positive in Unitization, we have:
:$x^{1/2} \le_A y^{1/2}$
{{qed}}