phuonghuyen123

$\sqrt[3]{a+3b}.1.1\leqslant \frac{1}{3}(a+3b+2)$

$\sqrt[3]{b+3c}\leq \frac{1}{3}.(b+3c+2)$

$\sqrt[3]{c+3a}\leq \frac{1}{3}.(c+3a+2)$

$\Leftrightarrow \sqrt[3]{a+3b}+\sqrt[3]{b+3c}+\sqrt[3]{c+3a}\leqslant \frac{1}{3}.(4a+4b+4c+6)=3$

    $''='' \Leftrightarrow$ $a=b=c=\frac{1}{3}$