Question

Cho a,b,c>0. CM:

\(2.\left(\frac{a}{b+2C}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)

Answer 1

trả lời

dùng bất đẳng thức cosi đc ko

hok tốt

Answer 2

undefined la gi

Answer 3

ta có

\(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\ge\frac{\left(a+b+c\right)^2}{3a+3b+3c}\ge\frac{a+b+c}{3}\)

\(\Leftrightarrow a=b=c=>\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}=1\)

tương tự 

\(\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\ge1\)

suy ra \(2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\ge2\)

=>\(1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\ge2\)

=> dpcm