Question

cho a,b,c>0. Chứng minh \(\dfrac{a+b}{c}\)+\(\dfrac{a+c}{b}\)+\(\dfrac{b+c}{a}\)≥ 6

Answer 1

Tham khảo:

Answer 2

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

\(=\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{a}{b}+\dfrac{c}{b}+\dfrac{b}{a}+\dfrac{c}{a}\)

\(=\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\)

Áp dụng BĐT cô si, ta có:

\(\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\)

\(\ge2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}+2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}+2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2+2+2=6\left(đpcm\right)\)