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?

  • 2
    $\begingroup$ 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 $x\neq y$) $\endgroup$ Feb 26, 2023 at 21:52
  • $\begingroup$ I'm guessing the best $f$ is $f(x)=\left|x-\frac12\right|.$ $\endgroup$ Feb 26, 2023 at 21:59
  • $\begingroup$ All you've shown us that $\int f\geq \frac16.$ But you haven't shown you can get $\frac16.$ $\endgroup$ Feb 26, 2023 at 22:01
  • $\begingroup$ The inequality implies $2f(x)\geq |2x-1|,$ so you certainly can't do better $\endgroup$ Feb 26, 2023 at 22:03
  • 1
    $\begingroup$ Solution on AoPS: artofproblemsolving.com/community/c7h1279762p6727193 – found with Approach0 $\endgroup$
    – Martin R
    Feb 27, 2023 at 5:56

1 Answer 1


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 $f(x) = |x-\frac{1}{2}|$.


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