Question

Cho a và b là các số dương, chứng tỏ :

                               \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)

Answer 1

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}-2\ge0\)

\(\Leftrightarrow\dfrac{a^2+b^2-2ab}{ab}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{ab}\ge0\left(ab>0\right)\)

Answer 2

Ta có: a,b > 0

=> \(\dfrac{a}{b},\dfrac{b}{a}>0\)

=> \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)