\(choA=\) \(2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)-\left(\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\right)\)
ta có : \(\frac{b}{b+2a}=1-\frac{2a}{b+2a}< =>-\frac{b}{b+2a}=\frac{2a}{b+2a}-1\)
tương tự ta có : \(-\frac{c}{c+2b}=\frac{2b}{c+2b}-1\)
\(-\frac{a}{a+2c}=\frac{2c}{a+2c}-1\)
\(=>A=2\left(\frac{a}{b+2c}+\frac{a}{a+2a}+\frac{b}{c+2a}+\frac{b}{c+2b}+\frac{c}{a+2b}+\frac{c}{a+2c}\right)-3\)(đoạn này mình làm hơi tắt nên bạn thông cảm nha)
Ta có: \(\frac{a}{b+2c}+\frac{a}{b+2a}+\frac{b}{c+2a}+\frac{b}{c+2b}+\frac{c}{a+2b}+\frac{c}{a+2c}\)
= \(\frac{a^2}{ab+2ca}+\frac{a^2}{ab+2a^2}+\frac{b^2}{bc+2ab}+\frac{b^2}{bc+2b^2}+\frac{c^2}{ca+2bc}+\frac{c^2}{ca+2c^2}\)
\(\ge\frac{\left(2\left(a+b+c\right)\right)^2}{2a^2+2b^2+2c^2+4ab+4bc+4ca}\)\(=\frac{4\left(a+b+c\right)^2}{2\left(a+b+c\right)^2}\)\(=2\)(sử dụng bđt SVACXO)
Vậy \(A\ge2.2-3=1\Rightarrow dpcm\)
dấu "=" xảy ra khi a=b=c
vậy A>= 1 khi a=b=c
