IMC / 1995 / Problems / Day 1, P2
Let fff be a continuous function on [0,1][0,1][0,1] such that for every x∈[0,1]x \in [0,1]x∈[0,1] we have ∫x1f(t) dt≥1−x22\int\limits_x^1 f(t)\,dt \ge \dfrac{1-x^2}{2}x∫1f(t)dt≥21−x2. Show that ∫01f2(t) dt≥13\int\limits_0^1 f^2(t)\,dt \ge \dfrac{1}{3}0∫1f2(t)dt≥31.
From the inequality 0≤∫01(f(x)−x)2dx=∫01f2(x) dx−2∫01xf(x) dx+∫01x2dx0 \le \int_0^1 (f(x) - x)^2 dx = \int_0^1 f^2(x)\,dx - 2 \int_0^1 x f(x)\,dx + \int_0^1 x^2 dx0≤∫01(f(x)−x)2dx=∫01f2(x)dx−2∫01xf(x)dx+∫01x2dx we get ∫01f2(x) dx≥2∫01xf(x) dx−∫01x2dx=2∫01xf(x) dx−13.\int_0^1 f^2(x)\,dx \ge 2 \int_0^1 x f(x)\,dx - \int_0^1 x^2 dx = 2 \int_0^1 x f(x)\,dx - \frac{1}{3}.∫01f2(x)dx≥2∫01xf(x)dx−∫01x2dx=2∫01xf(x)dx−31. From the hypotheses we have ∫01∫x1f(t) dt dx≥∫011−x22 dx\int\limits_0^1 \int\limits_x^1 f(t)\,dt\,dx \ge \int\limits_0^1 \frac{1-x^2}{2}\,dx0∫1x∫1f(t)dtdx≥0∫121−x2dx or ∫01tf(t) dt≥13\int\limits_0^1 t f(t)\,dt \ge \frac{1}{3}0∫1tf(t)dt≥31. This completes the proof.