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Sa se determine functia \( f:\mathbb{R}_{+}^{*} \to\mathbb{R} \) cu proprietatea ca

\( \ln(xyz)\leq xf(x)+yf(y)+zf(z)\leq xyz\cdot f(xyz), \forall x, y, z\in\mathbb{R}_{+}^{*} \).

R.M.I. C-ta, 2005

Intai luam \( x=y=z \) si obtinem

\( f(x)\geq \frac{\ln x}{x} \) \( (*) \) si \( 3f(x)\leq x^2f(x^3) \) pentru orice \( x\in \mathbb{R}^{*}_{+} \).

In aceste relatii daca luam \( x=1 \) obtinem \( f(1)\geq \ln1=0 \) si \( 3f(1)\leq f(1) \) adica \( f(1)\leq 0 \).

Deci obtinem \( f(1)=0 \).

Luam iar in relatia din enunt \( z=1 \) si avem

\( \ln(xy)\leq xf(x)+yf(y)\leq xy f(xy). \)

Luam \( y=\frac{1}{x} \) si avem

\( 0\leq xf(x)+\frac{1}{x} f(\frac{1}{x})\leq f(1)=0 \).

Deci \( xf(x)+\frac{1}{x} f(\frac{1}{x})=0 \) de unde rezulta \( \frac{1}{x} f(\frac{1}{x})=-xf(x) \).

In relatia \( (*) \) daca luam \( x \rightarrow \frac{1}{x} \) obtinem

\( \frac{1}{x} f(\frac{1}{x}) \geq \ln(\frac{1}{x}) \)

\( -xf(x)\geq -\ln x \)

\( xf(x)\leq \ln x \) \( (**) \).

Din relatiile \( (*) \) si \( (**) \) rezulta ca \( f(x)=\frac{\ln x}{x}. \)