Question

Cho a,b,c là các số thực dương. CMR: 

\(\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\ge4\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)

Answer 1

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

Ta có \(\left(\dfrac{b}{c}+\dfrac{b}{a}\right)\left(a+c\right)\ge\left(\sqrt{b}+\sqrt{b}\right)^2=4b\Leftrightarrow\dfrac{b}{c}+\dfrac{b}{a}\ge\dfrac{4b}{a+c}\)

CMTT \(\Leftrightarrow\left(\dfrac{c}{a}+\dfrac{c}{b}\right)\ge\dfrac{4c}{a+b};\dfrac{a}{b}+\dfrac{a}{c}\ge\dfrac{4a}{b+c}\)

Cộng VTV ta đc đpcm

Dấu \("="\Leftrightarrow a=b=c\)