# Math Competition Question for IMC 2016 (University level)

I am practicing for the IMC math competition for university students, and I was wondering if someone could help me with this question:

Today, Ivan the Confessor prefers continuous functions $$f:[0,1]\to\mathbb R$$ satisfying $$f(x)+f(y)\geq |x-y|$$ for all $$x,y\in [0,1]?$$ Find the minimum of $$\int_0^1 f$$ over all preferred functions.

This is from the 2016 IMC, question 7.

My idea was to integrate on both sides from 1 to 0 twice, the first integral in x and the second in y, but this gave a solution of 1/6, but the solution says the true answer is 1/4. I was wondering if someone could explain why my answer was incorrect, and if anyone has a solution to this problem?

• Your method assumes there is a function such that $f(x)+f(y) = |x-y|$ for all $x,y \in [0,1]$. This function cannot exist. (Check $x=y$, and then $x\neq y$) Feb 26, 2023 at 21:52
• I'm guessing the best $f$ is $f(x)=\left|x-\frac12\right|.$ Feb 26, 2023 at 21:59
• All you've shown us that $\int f\geq \frac16.$ But you haven't shown you can get $\frac16.$ Feb 26, 2023 at 22:01
• The inequality implies $2f(x)\geq |2x-1|,$ so you certainly can't do better Feb 26, 2023 at 22:03
• Solution on AoPS: artofproblemsolving.com/community/c7h1279762p6727193 – found with Approach0 Feb 27, 2023 at 5:56

## 1 Answer

Take $$y=1-x$$. Then $$f(x) + f(1-x) \geq |2x-1|$$, so $$\int_0^1 f(x) dx + \int_0^1 f(1-x)dx \geq \int_0^1 |2x-1| dx = \frac{1}{2}.$$

However, using the substitution $$u=1-x$$, one can see that $$\int_0^1 f(1-x) dx = \int_0^1 f(x) dx$$. Therefore $$\int_0^1 f(x) dx \geq \frac{1}{4}$$. The equality occurs for the ''preferred'' function $$f(x) = |x-\frac{1}{2}|$$.