2022 IMO Problem 2: Find all functions with given condition

2022 IMO Problem 2: Find all functions with given condition

Let R+ denote the set of positive real numbers. Find all functions f : R+ → R+ such that for each x ∈ R+, there is exactly one y ∈ R+ satisfying xf(y) + yf(x) less or equal to 2.

AM-GM Inequality and some intuitions are used in solving this problem.

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The steps are:

00:00:00 Intro to the Problem Statement: IMO 2022 - Problem 2
00:00:56 Examine the symmetry of the condition
00:02:46 The unique y value must be the same as x
00:06:39 Function f(x) is bounded above by 1/x
00:07:22 Function f(x) must be positive
00:17:51 Claim that f(x)=1/x and Summary

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[hensgen]

There are 2 (unfortunately negative) comments on this.

1) At 7:25, to prove that f(x)>0 is superfluous, since this is part of the hypothesis on f.

2) Worse, at 15:35 the "WTS" inequality σ/x0 - δσ/x0 < 0 is wrongly transformed into σ(1-δ) < x0; however, the WTS inequality is obviously equivalent to 1-δ < 0 which is wrong by definition of δ. As a remedy, the proof can be finished along these lines if one also takes the term (so far neglected) -σ/(x0+σ) into account.