Đặt $A=2(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}-(\dfrac{b}{b+2a}+\dfrac{c}{c+2b}+\dfrac{a}{a+2b}$
$=2(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b})+3-(\dfrac{b}{b+2a}+\dfrac{c}{c+2b}+\dfrac{a}{a+2b})-3$
$=2(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b})+1-\dfrac{b}{b+2a}+1-\dfrac{c}{c+2b}+1-\dfrac{a}{a+2b}-3$
$=(\dfrac{2a}{b+2c}+\dfrac{2b}{c+2a}+\dfrac{2c}{a+2b})+\dfrac{2a}{b+2a}+\dfrac{2b}{c+2b}+\dfrac{2c}{a+2c}-3$
$=2a(\dfrac{1}{b+2c}+\dfrac{1}{b+2a})+2b(\dfrac{1}{c+2a}+\dfrac{1}{c+2b})+2c(\dfrac{1}{a+2b}$
$+\dfrac{1}{a+2c})-3$
Áp dụng bđt ở câu a ta có:
$\dfrac{1}{b+2c}+\dfrac{1}{b+2a}≥\dfrac{4}{2(a+b+c)}=\dfrac{2}{a+b+c}$
$⇒2a(\dfrac{1}{b+2c}+\dfrac{1}{b+2a})≥\dfrac{2.2a}{a+b+c}=\dfrac{4a}{a+b+c}$
Chứng minh tương tự ta có:
$2b(\dfrac{1}{c+2a}+\dfrac{1}{c+2b})≥\dfrac{4b}{a+b+c}$
$2c(\dfrac{1}{a+2b}+\dfrac{1}{a+2c})≥\dfrac{4c}{a+b+c}$
$⇒2a(\dfrac{1}{b+2c}+\dfrac{1}{b+2a})+2b(\dfrac{1}{c+2a}+\dfrac{1}{c+2b})+2c(\dfrac{1}{a+2b}$
$+\dfrac{1}{a+2c})≥\dfrac{4a}{a+b+c}+\dfrac{4b}{a+b+c}+\dfrac{4c}{a+b+c}=$
$\dfrac{4(a+b+c)}{a+b+c}=4$
$⇒2a(\dfrac{1}{b+2c}+\dfrac{1}{b+2a})+2b(\dfrac{1}{c+2a}+\dfrac{1}{c+2b})+2c(\dfrac{1}{a+2b}$
$+\dfrac{1}{a+2c})-3≥4-3=1$
Hay $A=2(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b})-(\dfrac{b}{b+2a}+\dfrac{c}{c+2b}+\dfrac{a}{a+2b})≥1$
$⇒đpcm$
Dấu `=` xảy ra $⇔\dfrac{1}{b+2c}=\dfrac{1}{b+2a}$;
$\dfrac{1}{c+2a}=\dfrac{1}{c+2b}$;
$\dfrac{1}{a+2b}=\dfrac{1}{a+2c}$
$⇔a=b=c$