Solutions to olympiad problems are seldom written with clarity in mind — just look at forum posts in the Art of Problem Solving. The author makes jumps and skips a bunch of steps, expecting the reader to fill in the gaps.

Usually this is not much of a problem — the missing steps become obvious when you sit down and think about what’s going on with a pencil and some paper. But sometimes, this is not the case.

### The problem

One of the worst examples I’ve seen comes in the book *Inequalities, A Mathematical Olympiad Approach*. By all means, this is an excellent book. Anyways, here’s one of its easier problems — and you’re expected to solve it using the triangle inequality:

Prove that for all real numbers a and b,

### Attempt 1: Intuitive solution

It isn’t clear how the triangle inequality fits. If I weren’t required to use the triangle inequality, I might be tempted to do an intuitive, case-by-case argument.

Let’s visualize the absolute value of as the difference between the two numbers on a number line. Now we compare this distance with the distance after you take the absolute value of both of them, . If one of the numbers is positive and the other negative, we clearly have a smaller distance if we ‘reflect’ the negative one over. Of course, if they’re both positive, or they’re both negative, then nothing happens and the distances remain equal.

There, a simple, fairly clear argument. Now let’s see what the book says.

### The book’s solution

Flip to the end of the book, and find

Consider and , and apply the triangle inequality.

Huh. Perhaps if you are better versed than I am in the art of solving inequalities, you’ll understand what this solution is saying. But I, of course, had no idea.

Maybe try the substitution they suggest. I only see one place to possibly substitute for anything — and substituting gives . Now what? I don’t think I did it right — this doesn’t make any sense.

To be fair, I cheated a little bit in the first attempt: I didn’t use the triangle inequality. Fair enough — let’s solve it with the triangle inequality then and come back to see if the solution makes any sense now.

### Attempt 2: Triangle inequality solution

A standard corollary to the triangle inequality of two variables is the following:

Combine this with the two variables switched around:

Combine the two inequalities and we get the desired

Now let’s look at the solution again. Does it make sense? No, at no point here did we do any substitution. Clearly the authors were thinking of a different solution that happened to also use the triangle inequality. Whatever it was, I had no idea what the solution meant.

### The book’s solution, decrypted

Out of ideas and hardly apt to let the issue rest, I consulted help online at a math forum. And look — it turns out that *my solution was without a doubt the same solution* as the book’s intended solution!

What the author meant was this: considering that , we have from the triangle inequality. Then, moving the over we get .

After that, the steps I took above are left to the reader.

Perhaps I’m a bit thick-headed, but your solution can’t *possibly* be very clear if a reader has the exact *same* solution yet can’t even *recognize* your solution as the same solution. Come to think of it, if I couldn’t even recognize the solution, what chance is there of anybody being able to *follow* the solution — especially if they’re new to inequalities?

Almost *every* one of the one-sentence phrasings of this solution I could think of would be clearer and less puzzling than the solution the book gives me.