Ta có:
$\dfrac{2}{a+3b}+\dfrac{1}{b+3c}+\dfrac{4}{c+3a}\ge\dfrac{49}{2(a+3b)+(b+3c)+4(c+3a)}=\dfrac{7}{2a+b+c}$
$\dfrac{4}{a+3b}+\dfrac{2}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{49}{4(a+3b)+2(b+3c)+(c+3a)}=\dfrac{7}{a+2b+c}$
$\dfrac{1}{a+3b}+\dfrac{4}{b+3c}+\dfrac{2}{c+3a}\ge\dfrac{49}{(a+3b)+4(b+3c)+2(c+3a)}=\dfrac{7}{a+b+2c}$
Cộng 3 BĐT trên ta có đpcm.