Question

Với a , b > 0 chứng minh \(\dfrac{a+b}{2}\ge\sqrt{ab}\)

Answer 1

bất đẳng thức cô-si ?

\(\dfrac{a+b}{2}\ge\sqrt{ab}\\ < =>a+b\ge2\sqrt{ab}\\ < =>\left(a+b\right)^2\ge4ab\\ < =>a^2+2ab+b^2\ge4ab\\ < =>a^2-2ab+b^2\ge0\\ < =>\left(a-b\right)^2\ge0\left(đúng\right)\)

=> \(\dfrac{a+b}{2}\ge\sqrt{ab}\)

chúc may mắn

Answer 2

Ta có: \(a\ge0;b\ge0\Rightarrow a+b\ge0\Leftrightarrow a^2+2ab+b^2\ge2ab\)

\(\Leftrightarrow\left(a+b\right)^2\ge2ab\Leftrightarrow a+b\ge\sqrt{2}.\sqrt{ab}\)

\(\Leftrightarrow2.\dfrac{a+b}{2}\ge\sqrt{2}.\sqrt{ab}\); do \(2>\sqrt{2}\) nên \(\dfrac{a+b}{2}\ge\sqrt{ab}\)