Recall this Russian entrance exam problem… Let be real numbers. If you set equal to any of the numbers , or in the equation , then at least one of the other two numbers will be a root of the resulting quadratic equation. Prove that .
Solution. I would have wanted to see what the prospective MGU students were supposed to do with it, as it seems to me there is some deep structure here I am missing!
We need first to think and then to compute, always a good idea. Define
Written in this second way it is clear that is symmetric: . Now let us start substituting the conditions of the problem: we must have that either or (or both…).
Let us assume that and that and see what follows. Then since , we must have that . So if we assume that , we have that and . However, it is easy to show that if and , then , a contradiction to our assumption. [This is how you do it: write the two equations and ; subtract them and solve for in terms of and and then substitute this solution into, say, .]
So the conclusion is that if , we must have . Similarly, if we assumed that but , we would get a contradiction. Hence the only possibility is that and at the same time. In other words, and are just the two solutions of the quadratic equation .
So far so good. Now we need to remember that are real and that . Let us solve the quadratic . We have that
since . As both and must be real, the expression under the square root must be nonnegative. Note that if the expression under the square root is zero, which is not allowed, so this expression must be positive, which happens if . At the same time we must have that , solving which inequality [Do this!], we get (if then ). Combining the two conditions, we get that as required.
I am puzzled by the following point: we had to manipulate and algebraically to show that these two equalities imply . Is there a way to see that this has to be the case without manipulation? In other words, what is the structure of the function that forces this implication?