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Fie \( f: [0,\infty) \to \mathbb{R} \) o functie derivabila, convexa cu \( f(0)=0 \).

a) Sa se arate ca \( \int_{0}^{x}f(t)dt \leq f^{\prime}(x)x^2/2, \forall x \in [0,\infty) \)
b) Sa se determine toate functiile pentru care avem egalitate.

Concursul "Grigore Moisil" 2008, Problema 3

a) f is convex, so for any t in [0,x] we have \( xf(t)\leq tf(x) \).

\( x\int_{0}^{x}f(t)dt\leq \frac{x^2}{2}f(x) \)

For x>0 simplify and get \( \int_{0}^{x}f(t)dt\leq \frac{x}{2}f(x) \) (1)

But \( f(x)=\int_{0}^{x}f^{\prime}(t)dt \leq xf^{\prime}(x) \) (2)
since f' increasing.

(1) and (2) prove the inequality

b) If \( \int_{0}^{x}f(t)dt=\frac{x^2}{2}f^{\prime}(x) \), then LHS is derivable, so RHS is derivable.

Derivate the equality =>

\( x^2y^{{\prime}{\prime}}+2xy^{\prime}-2y=0 \)

\( f(x)=ax+\frac{b}{x^2} \)

\( f(0)=0 \)

\( f(x)=ax \)