= Solution
Compare the hull with the empty hull and the filled unit half-disc $J=\{z\in\mathbb H:|z|\leq1\}$. Their <mapping-out functions of compact H-hulls> are $g_\varnothing(z)=z$ and $g_J(z)=z+1/z$, the latter also giving the <half-plane capacity of a half-disc>.
Use one <planar Brownian motion> started at $iy$, with $y>1$. For $x>1$, exiting $\mathbb H\setminus J$ through $(x,\infty)$ entails avoiding $K$, while exiting $\mathbb H\setminus K$ through this interval entails reaching it before exit from the whole half-plane. Thus the tail <harmonic measures> satisfy
$$
p_J(y,x)\leq p_K(y,x)\leq p_\varnothing(y,x).
$$
For any of these fixed hulls $L$, the <Poisson kernel for the upper half-plane> gives the exact tail formula
$$
p_L(y,x)=\frac12-\frac1\pi
\arctan\!\frac{g_L(x)-\operatorname{Re}g_L(iy)}{\operatorname{Im}g_L(iy)}
=\frac12-\frac{g_L(x)}{\pi y}+o(y^{-1}).
$$
Subtract from $1/2$, multiply by $\pi y$, and take the limit. The inequalities reverse, yielding \b[the real boundary bounds]:
$$
\boxed{x\leq g_K(x)\leq x+\frac1x\qquad(x>1).}
$$
Applying this result to the reflected hull $-\overline K$ also gives
$$
x+\frac1x\leq g_K(x)\leq x\qquad(x<-1).
$$
The comparison is of Brownian exit events, so it does not assume that arbitrary conformal maps are pointwise ordered by domain inclusion.
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