Ta có:
$\dfrac{1}{\sqrt[3]{a+3b}}+\dfrac{1}{\sqrt[3]{b+3c}}+\dfrac{1}{\sqrt[3]{c+3a}}$
$=\dfrac{12}{3\sqrt[3]{8.8.(a+3b)}}+\dfrac{12}{3\sqrt[3]{8.8.(b+3c)}}+\dfrac{12}{3\sqrt[3]{8.8.(c+3a)}}$
$\ge\dfrac{12}{a+3b+16}+\dfrac{12}{b+3c+16}+\dfrac{12}{c+3a+16}$
$=\dfrac{12}{a+3b+16}+\dfrac{a+3b+16}{48}+\dfrac{12}{b+3c+16}+\dfrac{b+3c+16}{48}+\dfrac{12}{c+3a+16}+\dfrac{c+3a+16}{48}-\dfrac{a+b+c}{12}-1$
$\ge2\sqrt{\dfrac{12}{a+3b+16}.\dfrac{a+3b+16}{48}}+2\sqrt{\dfrac{12}{b+3c+16}.\dfrac{b+3c+16}{48}}+2\sqrt{\dfrac{12}{c+3a+16}.\dfrac{c+3a+16}{48}}-\dfrac{3}{2}=\dfrac{3}{2}$
Dấu bằng xảy ra khi và chỉ khi $a=b=c=2$.