Hm, I don't see, how you can make the Robertson uncertainty relation stronger, because the equal sign usually can be reached. For the position-momentum uncertainty relation these are the Gaussian wave packets. The most simple derivation of the uncertainty relation just uses the positive definiteness of the scalar product. We assume that the system is prepared in a pure state represented ##|\psi \rangle## (normalized to 1).
Assume that \langle A \rangle=\langle B \rangle (expectation values wrt. ##|\psi \rangle##). Otherwise consider the operators \hat{A}-\langle A \rangle \hat{1} and \hat{B}-\langle B \rangle \hat{1}.
Then we have to evaluate the standard deviations by
\langle \Delta A^2 \rangle=\langle \psi|\hat{A}^2|\psi \rangle, \quad \langle \Delta B^2 \rangle=\langle \psi|\hat{B}^2|\psi \rangle.
To that end we define the 2nd-order polynomial
$$P(\lambda)=\langle \psi|(\hat{A}-\mathrm{i} \lambda \hat{B})(\hat{A}+\mathrm{i} \lambda \hat{B})|\psi \rangle,$$
which is real, i.e., ##P(\lambda) \in \mathbb{R}## for ##\lambda \in \mathbb{R}##. To prove this, note that for ##\lambda \in \mathbb{R}## the operator in the brackets is self-adjoint and thus
$$P(\lambda)=\langle (\hat{A}+\mathrm{i} \lambda \hat{B}) \psi|(\hat{A}+\mathrm{i} \lambda \hat{B}) \psi \rangle, \quad \lambda \in \mathbb{R} \qquad (*).$$
Since the scalar product is positive definite this implies ##P(\lambda) \geq 0## for ##\lambda \in \mathbb{R}##. Multiplying out the operator square, you get
$$P(\lambda)=\lambda^2 \Delta B^2 + \Delta A^2 + \lambda \langle \psi|\mathrm{i} [\hat{A},\hat{B}]|\psi \geq 0, \quad \lambda \in \mathbb{R}.$$
Note that ##\mathrm{i} [\hat{A},\hat{B}]## is self-adjoint and thus all coefficients are real, which again proves that ##P(\lambda)## is a real polynomial.
The discrimant of this quadratic polynomial must thus be negative or 0 (otherwise you'd have 2 real zeros, and the polynomial could not be positive semidefinite). This implies
$$\Delta A^2 \Delta B^2 \geq \frac{1}{4} \langle \psi|\mathrm{i} [\hat{A},\hat{B}]|\psi\rangle^2$$
or
$$\Delta A \Delta B \geq \frac{1}{2} |\langle \psi|\mathrm{i} [\hat{A},\hat{B}]|\psi\rangle|.$$
This is the general Robertson-uncertainty relation.
It also shows, when the equality sign holds. That's only the case if the discriminant of the quadratic equation is 0, and then the polynomial has a double zero at some real ##\lambda##. Thus, due to (*) and the positive definiteness of the scalar product, the equality sign can hold if and only if there is a state ##|\psi \rangle##, for which
$$(\hat{A}+\mathrm{i} \lambda \hat{B}) |\psi \rangle=0$$
for some ##\lambda \in \mathbb{R}##.
[EDITED on 12/12/14 from here on:]
Let's take position and momentum as an example. This is most easily worked out in the position representation, where \hat{x}=x and ##\hat{p}=-\mathrm{i} \partial_x##, acting on the Hilbert space ##L^2(\mathbb{R})##. To get the most general case, we look for "minimal-uncertainty states", with general averages for both ##x## and ##p##, which we call ##\bar{x}## and ##\bar{p}##. Then from (*) and the discussion thereafter we know that the equality sign in the Robertson uncertainty relation is valid, if there exists a real ##\lambda## and normalizable ket ##|\psi \rangle##, which fulfills
$$[\lambda (\hat{p}- \mathrm{i} \bar{p} \hat{1})+(\hat{x}-\bar{x} \hat{1})]|\psi \rangle=0.$$
In position representation this reads
$$\lambda(\partial_x - \mathrm{i} \bar{p})+(x-\bar{x}) \psi(x)=0$$
or
$$\partial_x \psi(x)=\left [-\frac{x-\bar{x}}{\lambda} + \mathrm{i} \bar{p} \right ] \psi(x)$$
Separation of variables leads immediately to the solution
$$\psi(x)=A \exp \left [-\frac{(x-\bar{x})^2}{2 \lambda}+\mathrm{i} \bar{p} x \right ].$$
This Gaussian wave packet is a proper state for all ##\lambda>0##. It's normalized for
$$A=\left (\frac{1}{\pi \lambda} \right )^{1/4}.$$
