Yes, that's an extremely hard problem. It's not difficult to find the answer with a computer algebra system if you know how to look for it. However, proving that it's the smaller is essentially mechanical application of algorithms.
This was extracted from this paper:
http://ami.ektf.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf
The main ideas are:
1) Notice that the equation x/(y+z) + y/(x+z) + z/(x+y) = 4 represents a space of dimension 2. This is seen by noticing that, if you have a solution (x,y,z), you automatically obtain another one by transforming to (ka,kb,kc).
2) So, it should be possible to reduce this to a
projective curve.
3) Since you only care about integral solutions, you are actually looking for rational points in that projective curve. One thing about algebraic curves is that they sometimes turn out to be equivalent to each other (think about isomorphic graphs, for example).
4) In that case, the pretty name for what we are looking for is
birational equivalence. It essentially means that you can find a transformation that maps rational numbers to rational numbers in another curve, so that any operation you do on one of them gets translated to another.
5) It turns out that this projective curve is isomorphic to an
elliptic curve, the transformation can be found in equation (2.1) of the paper I linked. Keep in mind that although it's a complicated thing to find, today most mathematical software implements some good algorithms and can tell you if the curve is isomorphic to an elliptic one, and the rational transformation if it is.
6) Reducing the problem is a good thing, because it means it can be solved (but it's not easy). The nice thing about them is that their algebra has a simple structure, you can "add" two points in the curve to obtain a third one. The details can be seen
here.
7) If you find a rational point at the curve, you can understand the algebra by adding the point to itself. Say you find a point A. Then you can compute a sequence:
A -> A + A -> A + A + A -> A + A + A + A -> ...
It might happen that you eventually reach a cycle. In that case, we have what we call a finite group. It also might happen that it goes on forever and you have an infinite chain.
8) These properties are encoded in the so-called
rank of the elliptic curve. A rank of 0 means you have a finite number of rational points. A rank of 1 means you have besides those an infinite chain, a rank of 2 means you have two infinite chains and so on.
9) Now, for the particular elliptic curve we obtain from that equation, it turns out that its rank is 1. This is very complicated to prove, but computer programs today can instantly find rational points and compute the rank.
10) When you check that the rank is 1 it's simple to enumerate all possible solutions and check if they give you positive x, y and z. You first check the ones in the finite group . If none of them work, you go to the ones in the infinite chain. When you keep summing the point to itself you get bigger and bigger numbers, so at the smallest "multiple" of the point you get the smallest solution. In the paper, the authors prove that repeating this always works to get positive x, y and z, but the probability is very small, and the answer turns out to be very large.
11) If you look at table 2, you see that the smallest solution for the equation has 81 digits!
Don't be disappointed if you don't understand all the steps in the solution. It's a very hard problem and most of the time the only way to work with this is to use computers.
Elliptic curves are very complicated objects, they are used in cryptography because their algebra is simple to implement, but extremely tough to understand. To have a notion of the complexity, the solution of
Fermat's Last Theorem can be summarized as saying that every elliptic curve can be understood as a
modular curve, which is some sort of projection of a generalized complex plane.
It turns out that proving this
simple statement took a ridiculously long time and the proof is extremely technical. Today it's widely believed that every algebraic structure arising in problems like this are essentially the same mathematical object (automorphic forms) viewed in different ways (see
here). No one has any good idea on how to prove this, but if it's done, Millennium problems like the Riemann hypothesis and the Birch and Swinnenton-Dyer conjecture will all be simple exercises in that theory! That's how complicated that subject is, if you understand it, you basically solved 21st century math!
which we can rewrite as
Using the identity on itself multiple times, we get
As k goes to infinity, we get
The left hand side is of the form theta*sin(x)/x (where x=theta/3k) as x goes to 0, for which the limit is theta. Then plugging in theta=pi gives
and dividing by 4 gives you the identity you posted.
