[tex]f(x) = \dfrac{x^2+x+1-2x}{x^2+x+1}= 1 - \dfrac{2x}{x^2+x+1}\\ \\\\\text{Trebuie doar sa se arate ca functia e strict crescatoare in}\\\text{intervalul }[\sqrt[3]{2009},\sqrt[3]{2010}]\\ \\\\ f'(x) = \dfrac{-2(x^2+x+1)+2x(2x+1)}{(x^2+x+1)^2} \\ \\ f'(x) = \dfrac{2x^2-2}{(x^2+x+1)^2}\\ \\ f'(x) = 0 \Rightarrow 2x^2 - 2 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1\\ \\ \Rightarrow f'(x) > 0,\quad x\in(-\infty,-1]\cup[1,+\infty)\\ \\ \Rightarrow f(x)-\text{strict crescatoare pe }(-\infty,-1]\cup[1,+\infty)\\ \\[/tex]
[tex][\sqrt[3]{2009},\sqrt[3]{2010}] \subset (-\infty,-1]\cup[1,+\infty) \Rightarrow\\ \\ \Rightarrow f(x)-\text{strict crescatoare pe }[\sqrt[3]{2009},\sqrt[3]{2010}]\\ \\\sqrt[3]{2009} \leq \sqrt[3]{2010}\Rightarrow \boxed{f(\sqrt[3]{2009}) \leq f(\sqrt[3]{2010})}[/tex]
