Doubts about gravitational lens magnification A gravitational lens approximately follows a $1/r$-curve, meaning that rays closer to the lensing object get bent more than those away. My question is, wouldn't that shrink any image as opposed to magnify it? 
Consider, for example, the sketch below.

The top ray from the star (that reaches the target) has to come out at a lower angle or else it wouldn't be converged by the gravitational lens (it'd go outward as shown by the uppermost ray). This means that the top and bottom ray have to meet at a smaller angle at the target (which would make a smaller image). 
Even when you have an Einstein ring, the rings should be thinner than the object by the same reasoning. 
Under what circumstances can something get magnified? Is there a circumstance where the area is necessarily larger and brighter? 
 A: Your picture is correct: those rays do indeed get compressed. But the angular size of the star in the transverse direction gets bigger, and it makes the overall image bigger. This makes sense physically: if at perfect alignment the image forms a ring, as the source approaches the optical axis it should get stretched in one direction and compressed in the other. Note that an Einstein ring is indeed thinner than the source, as far as I can tell.
The geometry is a bit involved, but it turns out that if $\theta$ is the angle the rays make with the optical axis (the observer-lens line) at the observer, $\alpha$ is the deflection angle, and $d\gamma$ is the angular size of the source with no lensing, then
$$\frac{d\theta}{d\gamma} = \frac{1}{1-d\alpha/d\theta},$$
which, since $d\alpha/d\theta < 0$, is smaller than 1 as you said.
The other direction is trickier, but you can imagine two light rays coming of off the sides of the source. If there were no lens, they would keep moving on a straight line and meet at some angle. The lens not only deflects them towards the observer but also towards each other, so that they meet at a larger angle than they otherwise would.
