[tex]\it \dfrac{2x^2+2x+3}{x^2+x+1} = \dfrac{2x^2+2x+2+1}{x^2+x+1}= 2+ \dfrac{1}{x^2+x+1}\ \ \ \ \ (*)\\ \\ \\ x^2+x+1=x^2+x+ \dfrac{1}{4}+ \dfrac{3}{4}=\Big(x+ \dfrac{1}{2}\Big)^2+ \dfrac{3}{4}\geq \dfrac{3}{4} \Rightarrow \dfrac{1}{x^2+x+1}\leq \dfrac{4}{3}|_{+2}\Rightarrow\\ \\ \\ \Rightarrow 2+ \dfrac{1}{x^2+x+1}\leq \dfrac{10}{3} \stackrel{(*)}{\Longrightarrow}\ \dfrac{2x^2+2x+3}{x^2+x+1}\leq \dfrac{10}{3},\ \forall x\in\mathbb{R}[/tex]
[tex]\it Deci,\ \ inegalitatea \ \ \dfrac{2x^2+2x+3}{x^2+x+1}\leq a\ este\ adev\breve arat\breve a\ pentru\ orice\ x\in\mathbb{R},\\ \\ \\ dac\breve a\ a\geq \dfrac{10}{3},\ adic\breve a\ a\in \Big[ \dfrac{10}{3},\ \infty\Big)[/tex]
