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Fie \( f:[0,1]\to \mathbb{R} \) o functie continua astfel incat pentru orice \( x \in [0,1] \) avem
\( \int_x^1f(t){\rm d}t\geq \frac{1-x^2}{2} \).
Demonstrati ca \( \int_0^1f^2(t){\rm d}t\geq \frac{1}{3} \).

IMC 1995

Notam \( F(x)=\int_x^1f(t){\rm d}t \).Avem ca \( \int_x^1f(t){\rm d}t\geq \frac{1-x^2}{2} \Rightarrow \int_{0}^{1}(F(1)-F(x))dx\geq \int_{0}^{1}\frac{1-x^{2}}{2}dx \Rightarrow F(1)-\int_{0}^{1}(x^{\prime}F(x))dx\geq \frac{1}{3} \Rightarrow \int_{0}^{1}xf(x)dx \geq \frac{1}{3} \)
Din inegalitatea CBS avem :\( \int_{0}^{1}f^{2}(x)dx\int_{0}^{1}x^{2}dx\geq (\int_{0}^{1}xf(x)dx)^{2}\geq \frac{1}{9} \), de unde rezulta concluzia.