IMC 2016 · Day 2 · P7

The minimum of $\int_0^1 f$ is $\frac14$.

Applying the condition with $0 \le x \le \frac12$, $y = x + \frac12$ we get $$f(x) + f\bigl( x + \tfrac12 \bigr) \ge \tfrac12.$$ By integrating, $$\int_0^1 f(x)\,dx = \int_0^{1/2} \bigl( f(x) + f(x + \tfrac12) \bigr)\,dx \ge \int_0^{1/2} \tfrac12\,dx = \tfrac14.$$ On the other hand, the function $f(x) = \bigl| x - \frac12 \bigr|$ satisfies the conditions because $$|x - y| = \bigl| \bigl( x - \tfrac12 \bigr) + \bigl( \tfrac12 - y \bigr) \bigr| \le \bigl| x - \tfrac12 \bigr| + \bigl| \tfrac12 - y \bigr| = f(x) + f(y),$$ and establishes $$\int_0^1 f(x)\,dx = \int_0^{1/2} \bigl( \tfrac12 - x \bigr)\,dx + \int_{1/2}^1 \bigl( x - \tfrac12 \bigr)\,dx = \frac18 + \frac18 = \frac14.$$