= Solution
Let $c=x(1_A)\geq0$. Positivity on $a^*a$ implies positivity on every <Positive element of a C-star algebra>, by taking its positive <square root>. For a <self-adjoint C-star element> $h$, the <positive cone of a C-star algebra> gives
$$
-\|h\|1_A\leq h\leq\|h\|1_A.
$$
Both $x(\|h\|1_A+h)$ and $x(\|h\|1_A-h)$ are nonnegative real numbers. Their difference shows $x(h)$ is real, and the same inequalities give $|x(h)|\leq c\|h\|$.
For an arbitrary $a\in A$, write $a=h+ik$ with $h=(a+a^*)/2$ and $k=(a-a^*)/(2i)$ self-adjoint. Since the <involution> is isometric, $\|h\|,\|k\|\leq\|a\|$. Hence
$$
|x(a)|\leq|x(h)|+|x(k)|\leq2c\|a\|.
$$
Thus \b[every <positive functional on a C-star algebra> is bounded], with the preliminary bound $\|x\|\leq2x(1_A)$. This proof does not presuppose the <Cauchy–Schwarz inequality for positive C-star functionals>. It also gives $x(a^*)=\overline{x(a)}$, since $x(h)$ and $x(k)$ are real.
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