Inegalitate integrala via un polinom de interpolare

[Cezar Lupu]

W.24. Fie \( f:[0,1]\to\mathbb{R} \) o functie integrabila astfel incat
\( \int_0^1f(x)dx=\int_0^1xf(x)dx=\int_0^1x^2f(x)dx=1. \)
Sa se arate ca \( \int_0^1 f^{2}(x)dx\geq 9. \)

Joseph Wildt International Contest, 2005


Provocare: In numarul din octombrie 2006 al revistei American Mathematical Monthly a aparut urmatoarea problema, propusa de Peter Pal Dalyay, care constituie o generalizare a problemei de mai sus, precum si a problemei date la ONM in 2004 (vezi http://www.mateforum.ro/viewtopic.php?t=16) si anume:

Fie \( f:[0,1]\to\mathbb{R} \) o functie continua astfel incat sa avem
\( \int_0^1x^{k}f(x)dx=1 \) pentru orice \( 1\leq k\leq n-1 \).
Sa se arate ca \( \int_0^1f^{2}(x)dx\geq n^2 \).

[Doru Popovici]

Fie polinomul \( p(x)=30x^2-24x+3 \).
\( \int_{0}^{1} (f(x)-p(x))dx=\int_{0}^{1} x(f(x)-p(x))dx=\int_{0}^{1}x^2(f(x)-p(x))dx \)

=> \( \int_{0}^{1} p(x)(f(x)-p(x))dx=0 \)
\( 0\leq \int_{0}^{1} (f(x)-p(x))^2dx=\int_{0}^{1} (f^2(x) -p(x)f(x) -p(x)(f(x)-p(x))dx= \) \( \int_{0}^{1} f^2(x)dx - \int_{0}^{1} p(x)f(x) -\int_{0}^{1} p(x)(f(x)-p(x))dx \)

Dar \( \int_{0}^{1} p(x)(f(x)-p(x))dx=0 \) => \( 0\leq \int_{0}^{1} f^2(x)dx - \int_{0}^{1}p(x)f(x)= \)
\( \int_{0}^{1}f^2(x)dx- \int_{0}^{1}30x^2f(x)dx+\int_{0}^{1}24xf(x)dx-\int_{0}^{1}3f(x)dx \)

Dar cf ipotezei:
\( 0\leq \int_{0}^{1}f^2(x)dx -30+24-3 \) => \( \int_{0}^{1}f^2(x)dx\geq 9 \).

[Doru Popovici]

Fie \( p(x)=10x^2-8x+1 \).

Cf inegalitatii CBS:

\( \int_{0}^{1}f^2(x)dx\int_{0}^{1}p(x)dx\geq (\int_{0}^{1}f(x)p(x)dx)^2 \)
Din calcul rezulta \( \int_{0}^{1}p(x)dx=1 \), iar \( \int_{0}^{1}f(x)(10x^2-8x+1)dx= \) \( 10\int_{0}^{1}x^2f(x)dx - 8\int_{0}^{1}xf(x)dx + 1\int_{0}^{1}f(x)dx=10-8+1=3 \)
=> \( \int_{0}^{1}f^2(x)dx\geq 3^2=9 \).