Positivity of adjoint products from spectral positivity ($a^{\ast }a\ge 0$)

Define positivity spectrally: a self-adjoint element has nonnegative spectrum. First, addition preserves it. For $p,q\ge 0$ and $s=\Vert p\Vert +\Vert q\Vert$, functional calculus gives $\Vert s1-p-q\Vert \le s$, forcing every real spectral value of $p+q$ to be nonnegative. The cone is proper because a self-adjoint element with spectrum contained in $\{0\}$ has norm zero.

For $h=a^{\ast }a$, let $h_{-}=\max (-h,0)$ by the continuous functional calculus, and set $c=a{h}_{-}^{1/2}$. Then $c^{\ast }c=-h_{-}^2$ is nonpositive. The nonzero spectra of products in opposite orders imply that $c{c}^{\ast }$ is nonpositive too. Write $c=u+iv$ with $u,v$ self-adjoint. Then $c^{\ast }c+c{c}^{\ast }=2(u^2+v^2)$ is positive, since self-adjoint squares have nonnegative spectrum. It is also nonpositive, hence zero. It follows that $c^{\ast }c=-c{c}^{\ast }$ is both positive and nonpositive, hence zero. The C-star identity gives $c=0$, so $h_{-}^2=0$ and $h_{-}=0$. Thus $h\ge 0$, without assuming positivity of adjoint products in an earlier step.