Why do all elements above $\require{mhchem}\ce{Fe}$ not decay to $\ce{Fe}$? OK, so $\ce{Fe}$ is the most 'stable element'. As such, why do all elements above it not decay into $\ce{Fe}$? In all cases, would it not lead to an increase in binding energy and therefore energy been released, meaning it is energetically feasible, and should happen spontaneously (given enough time)? 
 A: There are a number of nuclei that can theoretically decay (based on conservation laws and energy) for which no decay has been observed.  A list is in Wikipedia.  There are more nuclei on it (164) than nuclei that are energetically prohibited from decay (90).  The lifetimes are long enough that the decays are not observed.
A: There are two separate issues to consider.
Firstly there is usually an energy barrier to decay. Radioactive decay occurs due to quantum tunnelling through the barrier, and the rate therefore depends on the barrier height. One of the very first studies of this was by George Gamow back in 1928, who studied the alpha decay of uranium-238. Even though alpha decay produces about 5Mev of energy (nearly 500 gigajoules per mole!!) the half life of uranium-238 is about the same as the age of the Solar System. Gamow's calculation is discussed in this PDF, or Google for many similar articles. The decay is slow because there is a barrier of around 25Mev that prevents the decay.
So while it may be energetically favourable for a nucleus to decay to iron a kinetic barrier may reduce the rate to a negligably small value.
Secondly, although for example nickel-60 may have a lower binding energy per nucleon than iron-56 $^1$ this does not mean the reaction:
$$ \mathrm{^{60}Ni \rightarrow {}^{56}Fe + \alpha }$$
is exothermic because the $\alpha$ particle also has a lower binding energy per nucleon than iron $^2$. If you took 56 nickel nuclei, disassembled them into individual nucleons then reassembled them into 60 iron nuclei you might get an overall energy decrease, but this route isn't available. Decay pathways are limited to $\alpha$, $\beta$ and fission, and if any step is not energetically favourable the decay process will stop at that step.
$^1$ Actually, according to Wikipedia nickel-62 is the most stable nucleus not iron-56
$^2$ I have no idea whether this reaction is exothermic or not
A: In general, you would need to consider cluster decay processes, which are extremely rare. You can estimate the decay probabilities of such processes using a formula given in this article.
