I’d like to discuss some of the problems given at this year’s International Mathematical Olympiad, held virtually in St. Petersburg, Russia. Let’s begin with the most controversial one – the inequality. By controversial I mean the most discussed one, with variety of opinions – from ardent supporters of this kind of problems to ardent opponents. I think, I know why, but let me first tell you that I like this inequality, it is somehow out of the box, and if this trend continues I might change my mind about whether inequalities should be given at math competitions. So, let me present some opinions from a popular math forum

A delightful problem, perhaps the best inequality in the history of IMO!

TelMarin

Not really. Just because is a nice problem it does not mean it is good for this competetion. And I can probably say this is the most inappropriate problem as a medium problem since it has solution which would be perfect as Undergraduate Olympiad

dangerousliri

Further, a guy, in order to back up the argument that it’s for Undergrad Olympiads, wrote

well you are not 15-17, are you? Cause when I first went to the IMO on 2018 I had not idea what concave or even continuous function even meant…

Hamel

Well, it’s not something to brag about that you went to IMO and had no idea what a continuous or concave function was, but well, I see the idea. By the way, in my country this stuff is taught in high school math program, so by no means it should be considered as undergraduate or rocket science. The only thing you need to solve this problem is that a concave function in some interval attains its minimum value either at or at And that a sum of concave functions (in some interval) is also concave. That’s it, and this stuff is taught in high school.

Along these lines, I don’t see why Karamata’s or Jensen’s inequalities, for example, are appropriate for IMO, but this one is not. On the contrary, this fact about concave/convex function is much more intuitive than some of the mentioned.

The issue apparently is that this problem is somewhat out of the box. Many people are ok to apply heavy lines of manipulating standard inequalities, but when it comes to some fresh thinking, they are frustrated.

Moreover, the mere idea is not a stroke of genius. So, you subtract the LHS from the RHS one and get a function You want to prove The first idea that comes into mind is to fix all the variables, except one, and to vary this one variable in order to see when gets its minimum. In this way, you may reduce the inequality to some special values of the variables. Sometimes, it is convenient to vary appropriately two of the variables and so on. In this particular case we shift all variables by some offset and calculate the value of the offset that makes minimal. I mean, this is not an unusual idea.

To be honest, the problem is not easy. I tried it, was pissed off at one point and saw the idea in the AoPS thread ([1], see post #7). I even considered the idea of shifting simultaneously all the variables, but somehow didn’t take into account (don’t know why) that the LHS didn’t change from that transformation. Ok, enough of this talk, here is the problem.

Problem 2, IMO 2021. Show that the inequality

holds for all real numbers

**Solution**. Denote the left hand side by the right hand side by and consider the function

We must prove The idea is to move (translate) the whole bunch of points and see when gets a minimum value. So, let Note that does not change by this transformation, that is, We have

Consider the function as a function of It is a concave in each of the intervals and Taking into account that sum of concave functions (in some interval) ia also concave, we get that is concave inside the intervals determined by the points

Now, any function that is concave in takes its minimum value in either when or when It means that attains its minimum value either when for some or when But since we obtained that the minimum value of is attained when for some . In case it yields and when we get

To recap, as we move the entire ensemble of points along the real axis, the minimum of is attained either when some hits or when some are symmetric about

In both cases, all the terms of that contain (respectively and ) are canceled no matter of the other values (resp. .) So, we remain with less variables and with a function with the same pattern. We do the same trick again and again and conclude that takes its minimum value when some values are zeroes and the others can be partitioned into symmetric pairs about In this case and the result follows.

**References**.

[1] IMO 2021, problem 2, AoPS thread.