Ta có:
\(\dfrac{1}{a+3b}+\dfrac{1}{c+3}\ge\dfrac{4}{a+3b+c+3}=\dfrac{4}{2b+6}=\dfrac{2}{b+3}\)
Tương tự:
\(\dfrac{1}{b+3c}+\dfrac{1}{a+3}\ge\dfrac{2}{c+3}\)
\(\dfrac{1}{c+3a}+\dfrac{1}{b+3}\ge\dfrac{2}{a+3}\)
Cộng vế:
\(\sum\dfrac{1}{a+3b}+\sum\dfrac{1}{a+3}\ge\sum\dfrac{2}{a+3}\)
\(\Rightarrow\sum\dfrac{1}{a+3b}\ge\sum\dfrac{1}{a+3}\) (đpcm)