You're assuming that nuclei with exactly a magic number of neutrons are more stable than all their non-magic neighbors in the chart of nuclides, but there's no reason to think that.
If a nucleus has a magic number of neutrons, that means one shell is completely full, and the next shell is empty. Therefore the next neutron you add (going to magic+1) will have significantly greater single-particle energy, so that nucleus should be less stable than the magic one. This is born out by both of your examples for the well-known magic number 126:
- Po-211 (N = 126+1): 0.5 seconds
- Ra-215 (N = 126+1): 1.5 milliseconds
However, if you remove one neutron, then nothing changes about the other (magic−1) neutrons; they are still in the same shells as in the magic case. No extra stability is caused by having a completely full shell as opposed to an almost full one. (In this respect, the word "magic" is misleading.)
Here's another way to think of it: In alpha decay (which is the main decay branch for all of your examples), two neutrons are removed from the nucleus. It will always be the two most energetic neutrons that are removed (those in the highest shells). In a magic nucleus, the lower shell is completely full, so two neutrons are removed from that lower shell. In a magic+1 nucleus, there is a lone neutron in the upper shell, so when two neutrons are removed, one comes from the upper shell and one from the lower shell. Since one comes from the upper shell, more energy is released in the alpha decay and in general the half-life will be shorter.
However, in a magic−1 nucleus, there are no neutrons in the upper shell, so both neutrons come from the lower shell. This is the same situation as the magic case, so there's no reason to expect the decay energies or half-lives to be drastically different. (Of course they won't be exactly the same, but the differences come from other, more subtle effects.)
BTW, 152 is not a "canonical" or "universal" magic number. It shows up on the plot you have there of some specific elements, but if you look at other elements, the gap between the shells occurs at a different place. 126 is a universal magic number, but at 152 the situation is more complicated because changing the neutron/proton ratio also shifts the shells relative to each other. This is why the thing I said about magic+1 always being less stable than magic doesn't hold for one of those. Nuclear structure is really complicated.
If protons decay, then what you say is true: all atomic nuclei are indeed unstable, and a so-called "stable" nucleus simply has too long a half-life for its decay to be observed.
The most tightly bound nucleus is $^{62}$Ni, with a binding energy per nucleon of 8.79 MeV [source], which is less than 1% of the mass of a nucleon. On the other hand, the decay of a proton through a process such as
$$p \to e^+ + \pi^0$$
results in the loss of most of the mass of the proton. So if the proton can decay then it's pretty clear that an atomic nucleus always has more much more mass than a hypothetical final state in which some or all of the protons have decayed. In other words, while neutrons do not decay inside "stable" atomic nuclei because of the binding energy of the nucleus, protons cannot be so protected because their decay would be much more energetically favourable (than that of a neutron to a proton).
The question of whether protons do decay is still unresolved, as far as I know.
If protons do not decay, then the $^1$H nucleus, by definition, is stable, so there is at least one stable nucleus.
Now, you might be wondering how we can establish that a nucleus is stable (assuming no proton decay). We make the assumption that energy is conserved, and it's impossible for a nucleus to be created if there isn't enough energy in the system to make up its rest mass. Given that assumption, say we have a nucleus. If we know the masses of the ground states of all nuclei with an equal or smaller number of nucleons, then we can rule out the possibility of there being a state that the given nucleus can transform into with less total mass. That in turn guarantees that the given nucleus is stable, since it can't decay into a final state with greater mass without violating conservation of energy. For a simple example, consider a deuteron, $^2$H. Its minimal possible decay products would be:
- a proton plus a neutron;
- two protons (plus an electron and an electron antineutrino)
- two neutrons (plus a positron and an electron neutrino)
- a diproton (plus an electron and an electron antineutrino)
- a dineutron (plus a positron and an electron neutrino)
But all of those states have higher mass than the deuteron, so the deuteron is stable; it has no decay channel.
Of course, you might wonder whether there are possible daughter nuclei whose masses we don't know because we've never observed them. Could, say, the "stable" $^{32}$S decay into $^{16}$P (with 15 protons and 1 neutron) and $^{16}$H (with 1 proton and 15 neutrons)? After all, we don't know the masses of these hypothetical nuclei. But if nuclei so far away from the drip line actually have masses low enough for that to happen, then there would have to be some radically new, unknown nuclear physics that would allow this to happen. Within anything remotely similar to existing models, this simply isn't possible.
Best Answer
This is really a comment, since I don't think there is an answer to your question, but it got a bit long to put in as a comment.
If you Google for "Why is technetium unstable" you'll find the question has been asked many times in different forums, but I've never seen a satisfactory answer. The problem is that nuclear structure is much more complex than electronic structure and there are few simple rules.
Actually the question isn't really "why is technetium unstable", but rather "why is technetium less stable than molybdenum and ruthenium", those being the major decay products. Presumably given enough computer time you could calculate the energies of these three nuclei, though whether that would really answer the "why" question is debatable.
Response to comment:
The two common (relatively) simple models of the nucleus are the liquid drop and the shell models. There is a reasonably basic description of the shell model here, and of the liquid drop model here (there's no special significance to this site other than after much Googling it seemed to give the best descriptions).
However if you look at the sction of this web site on beta decay, at the end of paragraph 14.19.2 you'll find the statement:
So these models fail to explain why no isotopes of Tc are stable, even though they generally work pretty well. This just shows how hard the problem is.