Two similar functional inequalities

(This is Glen.)

I was sorting through my stuff and found a piece of paper from when I tried last year's NTST problems, which I presumably saved because I thought I might eventually want to write a post about it. So here I am, writing a post so I can safely dump this piece of paper into my recycling bin.

One functional inequality

Here's the relevant problem:

(ISL 2023 A4)  Let $\mathbb R_{>0}$ be the set of positive real numbers. Determine all functions $f \colon \mathbb R_{>0} \to \mathbb R_{>0}$ such that \[x \left(f(x) + f(y)\right) \geqslant \left(f(f(x)) + y\right) f(y)\] for every $x, y \in \mathbb R_{>0}$.

This was NTST 2024 Day 1 Q2, by the way.

To avoid spoilers, I'm going to put a very crudely censored version of my working, and then reveal each bit one by one, in the order that I think I wrote everything in. Here we go:

Anyway, let's begin.

Rectangle #1

Yay, it's the problem statement!

Rectangle #2

Here, I tried some simple substitutions. These were generally motivated by the aim to make things cancel:

• Substituting $x=y$ lets the $xf(x)$ and $yf(y)$ terms cancel, which gives the nice-looking $x\ge f(f(x))$. That looked useful, so I boxed it.
• Substituting $x=f(y)$ produces a $f(f(y))$ on the LHS, which we now know is $\le f(y)$. Unfortunately, this just gives us $f(x)\ge f(f(f(x)))$, which we already know.
• Substituting $y=f(x)$ makes the two things in the brackets equal, so they cancel (it's nice not having to worry about cancelling zeroes), but again this gives something we already know.
• Replacing $y$ with $f(y)$ makes the stuff in the LHS bracket look like the stuff in the RHS bracket, so we can chain inequalities. However, this gives something that doesn't look very nice.

Rectangle #3

It seemed that just plain substitutions weren't going to get me something useful, so I decided to try something else. Notice that the LHS of the functional inequality is almost symmetric, so we can swap $x$ with $y$ and getting something similar. Adding this to the original inequality gives the thing on the top, and after cancellation, we get...something which is implied by $x\ge f(f(x))$.

Rectangle #4

Not entirely sure why I wrote this but I probably wanted to try something that I realised wouldn't work. Or maybe I was sending condolences to myself.

Rectangle #5

At this point, I realised that I didn't actually know what the solution set was, so I decided to guess what $f$ could be. $f(x) = \frac1x$ is a very common solution for $\mathbb{R}^+\rightarrow \mathbb{R}^+$ problems, and it worked for this. I then tried $f(x) = \frac{a}x$, where $a>0$, and that worked as well. That seemed like a reasonable family of solutions, so I figured that could likely be the answer.

Knowing this was incredibly useful and I really should have done it earlier: if you have a guess of what the solution set could look like, that can give you some intuition about what to try. For example, because $f(x)=\frac1x$ isn't defined at $0$, we may rule out continuity tricks that involve extending the domain of $f$ to $0$.

On the other hand, if we wanted to show that $f(x) = \frac{a}x$ for some $a$, that would be equivalent to showing that $xf(x)$ is constant, which does look like something that can be done from our expressions...

Rectangles #6 and #7

Conveniently, the problem statement has both an $xf(x)$ and a $yf(y)$. We want them to be equal, so let's move them to the same side. Now, if $yf(y) > xf(x)$, then this tells us something new: that $x > f(f(x))$. Progress!

Rectangle #8

Oops, this deduction is wrong. What is true is something slightly weaker: for each $x$, if $xf(x)$ isn't the maximum value (so there's some $y$ such that $yf(y) > xf(x)$), then $x > f(f(x))$.

I tried to derive a contradiction from something along these lines, but couldn't think of anything that worked.

Rectangle #9

I realised that there was a way to get $x - f(f(x))$ (which is $>0$ for some $x$ if $xf(x)$ isn't constant) on the smaller side of the inequality. This is a substitution that I did back in Rectangle #1 and forgot about, and gives that $f(x) - f(f(f(x))) \ge x - f(f(x))$.

But hold on: we can replace $x$ with $f(x)$, and then we have $f(f(x)) - f(f(f(f(x)))) \ge f(x) - f(f(f(x))) \ge x - f(f(x)) > 0$. What does this mean?

Rectangle #10

If we tried to plot $x, f(f(x)), f(f(f(f(x)))), \ldots$ on a number line (mine goes from bottom to top instead of from left to right), what would they look like? Well, $f(f(x))$ would be some distance below $x$. Then, $f(f(f(f(x))))$ would be some distance below $f(f(x))$, and this distance is at least the distance between $f(f(x))$ and $x$. Proceeding similarly, each term is at least $x - f(f(x))$ below the previous one, and hence we eventually go below $0$, contradiction! (The $\#$ is how people write the contradiction symbol in the UK. It's easier to write than $\rightarrow\leftarrow$, so I use that now.)

Summary

I've written quite a bit, but if you think about it, there wasn't very much to the solution. These were the main steps involved:

• Try some simple substitutions, in particlar, $x=y$ and $x=f(y)$.
• Suppose that $f$ isn't in our guessed solution set and obtain some (strict) inequality.
• Visualise what this means, and observe a contradiction.

All in all, this reminded me very strongly of a certain other recent functional inequality...

Another functional inequality

(IMO 2022/2) Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for each $x \in \mathbb{R}^+$, there is exactly one $y \in \mathbb{R}^+$ satisfying $$xf(y)+yf(x) \leq 2$$

There are some superficial similarities to the other problem: both are functional inequalities with $f:\mathbb{R}^+\rightarrow \mathbb{R}^+$. But as it turns out, the solution for this also goes along similar lines. (More accurately, I remember having a solution that proceeded along similar lines. I had to solve this again in order to write this post and it might have been a self-fulfilling prophecy that I found something that fits to the framework above.) If you haven't done this problem before, you might want to try it out before reading my solution below.

Step 0: Guess the solution set

Earlier, I made the mistake of running into the problem blindly without any idea what the solution set could be. That's not very advisable. Let's try to do things right this time. Anyway, the usual guess of $f(x)=x$ fails, but $f(x)=\frac1x$ works by AM-GM (and in this case, the unique $y$ is just $y=x$). Variations of this end up not working, and so we might hypothesise that this is the only solution.

Step 1: Simple substitutions observations

Well, we can't exactly substitute stuff, because we don't have an (in)equality to work with. However, we can make some observations. For example, notice that the symmetry of $xf(y)+yf(x) \leq 2$ means that if $y$ is the unique value for $x$, then $x$ must play the corresponding role for $y$. In other words, every $x$ is either paired with itself (so $xf(x) \le 1$), or must be paired (monogamously) with some other $y$.

Step 2: Obtain a strict inequality

In our (proposed) solution set, every $x$ is paired with itself. What if some $x$ didn't satisfy this? Then $xf(x)>1$, and its partner $y$ must also satisfy $yf(y)>1$. Great, that means that $xf(y) + yf(x) \le 2 < xf(x) + yf(y)$, which rearranges to $$(x-y)(f(x)-f(y)) > 0.$$ That seems useful. WLOG $x<y$, then $f(x)<f(y)$, but now $xf(y)+yf(x) > xf(x) + xf(x) >2$, contradiction.

(Edit: I just had a look at AoPS and this step can be done faster: just note that $xf(y) + yf(x) > \frac{x}y + \frac{y}x \ge 2$ by AM-GM.)

In other words, $xf(x) \le 1$ for each $x$.

Step 2, again

Let's be more ambitious: suppose that there's some solution that isn't $f(x) = \frac1x$, then we must have $f(x) < \frac1x$, a strict inequality. Can we try to abuse this?

Step 3: Visualise

Intuitively, because $f(x)$ is now "too small", there is probably some other $y\ne x$ such that $xf(y) + yf(x) \le 2$. Let's try to force this to happen. Note that $xf(y) + yf(x) \le \frac{x}y + yf(x) < \frac{x}y + \frac{y}x$ for each $y$. Let's graph the two things on the right as $y$ varies:

By AM-GM, the top graph has a minimum point of $(x,2)$. Since the other graph sits beneath it, it has to go below the value $2$.

But hold on: this means that this graph is below $2$ for a whole range of $y$, so there's some $y\ne x$ that's near $x$ for which $\frac{x}y + yf(x) < 2$, contradiction.

(What we are really using is the fact that $\frac{x}y + yf(x)$ is continuous in $y$: to make this rigorous, we'd have to apply the $\varepsilon-\delta$ definition of continuity to $\frac{x}y+yf(x)$. Alternatively, we could avoid quoting continuity with the "bare hands" approach of putting in $y=x+\varepsilon$ and showing algebraically that the expression is $<2$ for small $\varepsilon$. Again, see Pengchong's post.)

Conclusion

Maybe the link between the two problems is more tenuous than it is in my head, but they just feel similar. But I do believe that my experience of solving IMO 2022/2 did, however subconsciously, guide me towards a solution of the NTST problem. This is, I guess, what happens to your brain after you've done enough Olympiad problems: problems start reminding you of other problems. Once in a while, this might lead you astray (see: IMO 2024/5 and overtraining), but most of the time (see: various posts in this blog, e.g. this, this, this or this) it turns out to be very helpful.