[tex]a\ \textgreater \ 0 \\ \\ a+\dfrac{1}{a}\geq2 \\ \\ $Conform inegalitatii mediilor: \quad \dfrac{x+y}{2} \geq \sqrt{x\cdot y},\quad \forall $ $x,y \ \textgreater \ 0 \\ \\ $Noi avem: \quad \dfrac{a+\dfrac{1}{a}}{2}\geq \sqrt{a\cdot \dfrac{1}{a}} \Rightarrow a+\dfrac{1}{a}\geq 2\cdot \sqrt{1}\Rightarrow \boxed{a+\dfrac{1}{a}\geq2}$ (A)[/tex]
