Best example of energy-entropy competition? What are the best examples in practical life
of an energy-entropy competition which favors entropy over energy?
My initial thought is a clogged drain -- too unlikely for the
hair/spaghetti to align itself along the pipe -- but this is probably
far from an optimal example.  Curious to see what you got.  Thanks.
 A: The air in this room.  Because of gravity, the lowest energy configuration is clearly one in which all the molecules lie on the floor.  But from the entropic point of view they should be exploring all of their phase space and bouncing around the room.
I would like to say also that examples such as the clogged drain and so on should be viewed purely as analogies.  Entropy is a real physical quantity that can be calculated.  As an actual physics problem, the calculation of the entropy of the drain + hair + water system would be a monster and there is no telling what the result will be! Is it even in equilibrium?
The Brazil nut effect cited above is a non-equilibrium effect, so although it is interesting I don't think it necessarily means anything about energy-entropy competition.  The balance of energy and entropy happens when a system (at constant temperature for example) reaches equilibrium, minimizing its free energy.  The shaken Brazil nuts are a non-equilibrium problem, and thus are not minimizing their free energy! But very interesting nevertheless, all the more so since it is non-equilibrium.
A: One example I know is so-called Brazil nut effect. When you place balls in a container of two different sizes and shake it, the larger ones will go up (even if they are denser than the smaller ones). So the final energy of system after introducing noise is clearly greater than the initial. I believe that the phenomena needs to be entropy-driven. However, I don't know a proof. 
A: Blackbody radiation: anything hotter than its environment radiates energy thus increasing the entropy of the universe. Entropy wins :-)
The Sun :-) The Sun's energy does not increase the Earth's total energy!
In fact, the Earth radiates almost exactly the same amount of energy as it receives from the Sun. What we really gain from the Sun is that we use the sun rays' low entropy to power life on Earth, and the Earth radiates high entropy microwaves in the night.
A: One of the nicest examples I know is the Kosterlitz-Thouless phase transition in the XY model. What is cool is that the transition is driven by the condensation of vortices which have an energy that diverges logarithmically with the size of the system. You would think they couldn't contribute at all because of this, but it turns out their entropy also diverges in the same way, so the free energy $F=E-TS \propto (c-k_BT) log(R)$ where $c$ is a parameter and $R$ is the size of the system. At sufficiently large $T$ the entropy terms wins and the system undergoes a transition through formation of vortices. 
p.s. after rereading the question I realize my answer does not involve "practical life" but I'll leave it anyway.
A: Well, as a physicist working mainly in the field of statistical physics, I obviously have to mention Ising model. In this case the model is actually tractable (at least in two dimensions) and can tell us a whole lot about (not just) energy-entropy battle. It's obvious that the ground state (for ferromagnetic case) is all spins pointing one way (say up). Now, if you point some spins the other way then you are losing in terms of energy (something like number of neighbors times number of wrong spins) but actually you gain hugely in entropy (because of translational invariance of lattice models). This argument can be made very precise when working with polymer model (which is isomorphic to Ising model) and considering low-temperature cluster expansion.

I am sorry, but I am not really able to provide nice references. Wikipedia articles are pretty bad. Maybe I should spend some time bringing them up to the current knowledge (that is to say, knowledge since like 1970s) about cluster expansion. For now, if anyone is interested, just read the basic paper on the topic.
A: If you are willing to go down to microscopic scales, a nice example of "entropy winning" is the phenomenon of depletion forces. Large particles in a suspension of smaller ones feel an effective attractive force, even if the interaction between all particles is just hard-wall. The attractive force arises because the volume available to the smaller particles increases when the larger ones get sufficiently close, and hence their entropy increases. See e.g.

Sho Asakura and Fumio Oosawa,
  "Interaction between particles suspended in solutions of macromolecules",
  J. Pol. Sci. 33 (1958) 183-192.

Depletion forces can be measured directly and are quite important for biological systems.
A: In my view it's not really correct to talk about an "energy-entropy competition".  I guess the reason people tend to think in those terms is because physicists like to speak of systems at a constant temperature minimising the Helmholtz free energy $A=U - TS$.  This looks like it's an energy term ($U$) added to a term involving the entropy $S$, with the temperature acting as a kind of conversion factor.  So it's easy to think that the system must be trying to minimise its energy and maximise its entropy at the same time, putting both factors in competition with each other.
But really that's not what's happening at all.  What's happening is that the system we're interested in is exchanging energy with a heat bath in order to remain at a constant temperature, and the entropy of the combined system (system+heat bath) is increasing towards a maximum.  If an amount $\Delta U$ of energy is transferred from the heat bath to the system of interest, causing a change $\Delta S$ in the entropy of the system of interest, then the total entropy has changed by $\Delta S - \Delta U/T$.
Now if we multiply this by $-T$ we get $\Delta U - T\Delta S$, which is equal to $\Delta A$, the change in Helmholtz free energy. Since the total entropy must be maximised, and since $T$ is assumed constant, $A$ must therefore be minimised. But now we can see that not only is the $TS$ term an entropy, but so is the $U$ one. The change in $U$ just represents the change in the entropy of the heat bath - it's just that multiplying it by $T$ has obscured that.  The competition is not between the system's energy and its entropy, but between the system's entropy and the heat bath's entropy.  The reason for multiplying by $-T$ is purely historical and has always struck me as rather unhelpful. Apart from anything else it's only permissible if $T$ is constant, whereas the $\Delta S - \Delta U/T$ formula works even if it isn't.
Having said all that I can now try to answer the question.  What we need is an example of a system where the $TS$ term dominates over the $U$ one.  One way to do this is just to prevent any heat from being transferred between the system and the heat bath.  For any system that's completely isolated from its environment, the $\Delta U/T$ term will go to zero and only the $\Delta S$ term will be left. So any chemical reaction or other physical process performed in a thermally insulated container will be entirely dominated by the entropy term.
But perhaps a more satisfying example would be an endothermic chemical reaction such as cooking an egg. In this case the $\Delta U$ term is negative, so the system actually sucks in heat from its environment, reducing its entropy. This is offset by a greater increase in the entropy of the system - so we can say the entropy of the system has "won" over the entropy of the heat bath.  Or in free energy terms, the $TS$ term has "won" over the $U$ one.
