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Sa se arate ca pentru orice functie continua \( f:[0,1]\rightarrow\mathbb{R} \) avem:

\( \frac{1}{4}\int_0^1f^2(x)dx+2\left(\int_0^1f(x)dx\right)^2\geq3\int_0^1f(x)dx\cdot\int_0^1xf(x)dx \).

Cezar Lupu SHL 2007

Sa dam o mica indicatie:

Scrieti inegalitatea din enunt astfel:

\( \int_0^1f^{2}(x)dx\geq 4\left(\int_0^1f(x)dx\right)\left(3\int_0^1xf(x)dx-2\int_0^1f(x)dx\right) \) si folositi inegalitatea elementara \( (x+y)^2\geq 4xy \).

Continuare la indicatie.

\( RHS\leq\left(\int_0^1{f(x)dx}+3\int_0^1{xf(x)dx}-2\int_0^1{f(x)dx}\right)^2=\left(\int_0^1{(3x-1)f(x)dx}\right)^2 \)

Solutia 2 (putin mai ,,muncitoreasca'')

Caut \( a\in\mathbb{R} \) astfel incat

\( \frac{1}{4}\int_0^1f^2(x)dx+(2+3a)\left(\int_0^1f(x)dx\right)^2\geq 3\int_0^1f(x)dx\int_0^1(x+a)f(x)dx \).

Notind \( y=\int_0^1f(x)dx \) e suficient sa gasim a astfel incat trinomul

\( (2+3a)y^2-3y\int_0^1(x+a)f(x)dx+\frac{1}{4}\int_0^1f^2(x)dx\geq0 \)

sa aiba \( \Delta \) nepozitiv si (2+3a) pozitiv.

Calculind \( \Delta=9\left(\int_0^1(x+a)^2f(x)dx\right)-(2+3a)\int_0^1f^2(x)dx\leq \)
\( \leq 9\int_0^1(x+a)^2\int_0^1f^2(x)dx-(2+3a)\int_0^1f^2(x)dx=(1+3a)^2\int_0^1f^2(x)dx \).

Asadar alegand \( a=-\frac{1}{3} \) obtinem inegalitatea dorita, cu egalitate daca \( f(x)=\lambda(x-\frac{1}{3}) \).