= Solution
A <compact H-hull> is a bounded relatively closed set $K\subset\mathbb H$ whose complement $D=\mathbb H\setminus K$ is a <simply connected domain>. Equivalently one can retain its compact closure in $\overline{\mathbb H}$; real <boundary> points do not affect the <mapping-out function>. The <Riemann mapping theorem> and <Schwarz reflection principle> near infinity give a unique <conformal map> $g_K:D\to\mathbb H$ with <hydrodynamic normalization at infinity>. Its <Laurent series> is
$$
g_K(z)=z+\frac{a_K}{z}+O(|z|^{-2}),\qquad a_K\in\mathbb R.
$$
The <half-plane capacity> is \b[$\operatorname{hcap}(K)=a_K$]. The missing constant term is part of the normalization, not a choice to be made again when comparing different hulls.
Here are the <conformal map> facts we use to justify positivity and comparison. The inverse $f_K:\mathbb H\to D\subset\mathbb H$ has expansion $f_K(w)=w-a_K/w+O(w^{-2})$. The upper-half-plane <Schwarz-Pick theorem>, applied to $w$ and $iY$ and then letting $Y\to\infty$, implies $\operatorname{Im}f_K(w)\geq\operatorname{Im}w$. Indeed the inequality between pseudohyperbolic distances gives
$$
\frac{4\operatorname{Im}f_K(w)\operatorname{Im}f_K(iY)}{|f_K(w)-\overline{f_K(iY)}|^2}
\geq\frac{4Y\operatorname{Im}w}{|w+iY|^2},
$$
and multiplying by $Y$ gives the height comparison in the limit. At $w=iY$, the inverse expansion reads $\operatorname{Im}f_K(iY)=Y+a_K/Y+O(Y^{-2})$, so $a_K\geq0$.
For $r>0$, uniqueness of the <mapping-out function> gives
$$
g_{rK}(z)=r g_K(z/r)
=z+\frac{r^2a_K}{z}+O(z^{-2}).
$$
Consequently
$$
\boxed{\operatorname{hcap}(rK)=r^2\operatorname{hcap}(K).}
$$
Translation by a real number $b$ similarly gives $g_{K+b}(z)=b+g_K(z-b)$ and leaves <half-plane capacity> unchanged.
If $K\subset A$ are <compact H-hulls>, their <mapping-out functions> compose as $g_A=g_L\circ g_K$, where $L$ is the filled image of $A\setminus K$. Comparing <Laurent series> yields
$$
\operatorname{hcap}(A)=\operatorname{hcap}(K)+\operatorname{hcap}(L)\geq\operatorname{hcap}(K).
$$
This proves <monotonicity of half-plane capacity>, rather than assuming it from inclusion alone. The filled half-disc $A=\{z\in\mathbb H:|z-b|\leq R\}$ has <mapping-out function>
$$
g_A(z)=z+\frac{R^2}{z-b},
$$
which takes its semicircular <boundary> to the real line and is a <conformal bijection> onto $\mathbb H$. Thus its <half-plane capacity> is $R^2$. Enclose $K$ in such a half-disc, compare capacities, and minimize over real centres and radii. This proves the sharp bound
$$
\boxed{\operatorname{hcap}(K)\leq\operatorname{rad}(K)^2.}
$$
If one defines the radius by an infimum, take $R$ strictly larger than that infimum and then let $R$ decrease; no attainment assumption is needed.
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