IMO 2022 P2 Review (Norway)| A Functional Inequality?

#IMO​ #2022 #Problem2 #Algebra #functional_inequality
In this video we review this year's IMO 2022, which was held in Norway, we solve Problem 2, which is a very cool functional inequality.
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[@Simio_Da_Tundra]

A simpler way to wrap things up:
We know that f(x)≤1/x. Suppose that for some x f(x)0 such that f(x)=1/(x+ε). Then, we know that y=x works (because 2xf(x)=2x/(x+ε)

[@fergalmdaly]

Thanks for the solution. The limit argument is very elegant but was surprising to me. When I did this I got the same sequence of results but for the last step, I thought "if it's < 1/x at some point x and x~y for some y!=x then that's going to force f(y) to be > 1/y somewhere near there". Instead of taking that limit, you look at it as a quadratic. The rule requires it to be > 2. You need to throw out the term that is known to be negative and then you can show that the remaining quadratic has a 2 roots and that one of them is > 0 and therefore a region in R+ where it goes < 0, contradicting the rule.
It's not as elegant as the limit but the idea that pushing f(x) away from 1/x in one spot is going to require it pushed in the opposite direction somewhere else is intuitive and applicable to other problems like this.

[@littlefermat]

Nice!
Indeed the limit and defining g is not the only way ;-)

[@fergalmdaly]

@@littlefermat actually I defined g. In fact I defined g at the start, which actually makes it harder to prove the earlier facts 😂

[@sithlordbinks]

Wow... very nice thank you little fermat sir!!!

[@deltaa-x4940]

PLEASE WE NEED THAT IN THE ARABIC CHANEL

[@deltaa-x4940]

Sir, when we get f(x+1)-f(x)=1
For f:Q--->Q
What's the which we can use to prove that f is linear

[@sithlordbinks]

Well this means that when x increments by 1, f(x) increments by 1, for every x, so use induction