Does anyone know of a good method for finding a lower bound of the Hausdorff dimension of a set $G$?
The only method I could find is to find an $\alpha$-Hölder function $f \colon G \to H$ then $\dim_H(G) \geq \alpha \dim_H(\operatorname{im}(f))$. Choosing $f$ cleverly will mean that $\operatorname{im}(f)$ will be a set whose Hausdorff dimension is already known (or at least a lower bound for it is known).
Upper bound for the Hausdorff dimension is often easy, from the definition.
Lower bound can be harder. One method can be used if you have a measure on your set. Even better, a measure that naturally fits with the structure of the set. Then lower bounds for the Hausdorff dimension come from density computations for that measure.
(Would citing my own book here be considered crass?)
Packing dimension may be opposite. The lower bound is easy from the definition, but the upper bound harder. Again a density with respeact to a measure can help with this upper bound.
added March 4
This density theorem is found in: G. Edgar, Integral, Probability, and Fractal Measures (Springer 1998) Theorem 1.5.14, p. 52.
Definitions ... Let $E \subseteq \mathbb{R}^n$ be a Borel set, let $\mu$ be a nonzero measure on $E$, let $s>0$ be a real number. Write $$ B_r(x) = \{y \in E \colon |y-x|\le r\} $$ for a closed ball. The upper $s$-density of $\mu$ at a point $x \in \mathbb{R}^n$ is $$ \overline{D}^s_\mu(x) = \limsup_{r \to 0} \frac{\mu(B_r(x))}{(2r)^s} . $$
A consequence of Theorem 1.5.14 is then: If $\sup_{x \in E} \overline{D}^s_\mu(x) < \infty$, then the Hausdorff dimension of $E$ is at least $s$.
There are lower bound based on potential theoretic methods. Still, you need to have a measure $\mu$ supported on your set $E$. Define the $s$-energy of $\mu$ as
$I_s(\mu) = \iint |x-y|^{-s} \mu({\rm d} x) \mu({\rm d} y)$
If $I_s(\mu) < \infty$, then $\dim_{\rm H} (E) \geq s$. [Theorem 4.13, Falconer Fractal Geometry 2nd Ed.] IIRC, the proof follows from the density-based lower bound that Gerald gave. But if your measure allows easy estimate of potential, then it might be more convenient.
Moreover, you have the following characterization of Hausdorff dimension:
$\dim_{\rm H} (E) = \inf \{ s \geq 0 : C_s(E) = 0 \} = \sup \{s \geq 0 : C_s(E) > 0\}$,
where $C_s(E)$ is the $s$-capacity of $E$, defined as
$C_s(E) = \sup\{I_s(\mu)^{-1}: \mu \text{ is a probability measure supported on }E \}$
There are also lower bounds based on the Fourier transform of $\mu$. See Sec. 4.4 of Falconer's book.
Finding good lower bounds of the Hausdorff dimension of the attractors to dissipative dynamical systems is a very difficult problem.
In the early 1970s Arnold explicitly stated it as an important open problem in case of the 2D Navier-Stokes equations in a bounded domain (see the compendium of Arnold's problems); in various forms the question probably goes back to Kolmogorov's seminar in the 1950s. If atrractor's dimension can be shown to grow indefinitely along the Reynolds number Re, this can be interpreted as a manifestation of the turbulence of the fluid flow.
The upper estimates of attractors' Hausdorff dimension are much easier. For instance, the best known result for the 2D Navier-Stokes equations with the Dirichlet boundary conditions in the domain $\Omega$ says that
$$\dim_H {\rm Attr}\leq \frac{|\Omega|}{\pi \nu^2}\ \|f\|_{L^2}$$
(here $\nu$ is the viscosity and $f$ is the external force). There are no satisfactory lower bounds of the Hausdorff dimension in this case. The situation is slightly better in case of the Navier-Stokes equations on a torus: good lower bounds are known but only for very specific forces $f$.
For one family of fractals there is a fairly straightforward method: If the set $G$ is the fixed point set of a number of similitudes $S_1,\ldots,S_n$ with contraction factors $s_1,\ldots,s_n$ then the Hausdorff dimension $D$ satisfies $$\sum_{i=1}^n s_i^D=1.$$ A bit more explanation may be in order: Each $S_i$ is an affine map on the form $S_i=a_i+s_iR_i$ where $R_i$ is a rotation (mirror symmetries allowed). These produce a map $S$ on the set of nonempty compact subsets of $\mathbb{R}^d$, say, given by $$S(F)=\bigcup_{i=1}^n S_i(F).$$ This map is a contraction in the Hausdorff metric on the space, and $G$ is the unique fixed point of the map.
There is one caveat: The sets $S_i(G)$ must be mutually disjoint, or almost so. For example the Sierpinski triangle: It is generated by three similitudes each mapping a given triangle to one half as big, so we should get $3\cdot(1/2)^D=1$. But the three subtriangles meet in three different points; however, these points are corners, and this counts as negligible.
I apologize for not having a reference handy. I learned this from a handwritten note which likewise lacked references, and I haven't had the fortitude to go and chase one down. If you can read Norwegian, there is a proof in my small note here. The main idea of the proof is fairly simple, however: It's a question of covering $G$ by small balls and seeing how these covers scale under $S_i$.
