Question

Cho a ≥3;tìm GTNN của S=a+1/a

Answer 1

\(S=a+\dfrac{1}{a}=\dfrac{a}{9}+\dfrac{8a}{9}+\dfrac{1}{a}\ge2\sqrt{\dfrac{a}{9}\cdot\dfrac{1}{a}}+\dfrac{8a}{9}=2\cdot\dfrac{1}{3}+\dfrac{8a}{9}\ge\dfrac{2}{3}+\dfrac{8\cdot3}{9}=\dfrac{2}{3}+\dfrac{8}{3}=\dfrac{10}{3}\\ S_{min}=\dfrac{10}{3}\Leftrightarrow a^2=9\Leftrightarrow a=3\)