Giải thích các bước giải:
Ta có :
$\sqrt[3]{a+3b}=\sqrt[3]{a+3b}.1.1\le \dfrac{a+3b+1+1}3=\dfrac{a+3b+2}3$
$\rightarrow \dfrac{1}{\sqrt[3]{a+3b}}\ge \dfrac{3}{a+3b+2}$
Tương tự ta có :
$ \dfrac{1}{\sqrt[3]{b+3c}}\ge \dfrac{3}{b+3c+2}$
$ \dfrac{1}{\sqrt[3]{c+3a}}\ge \dfrac{3}{c+3a+2}$
$\rightarrow P\ge\dfrac{3}{a+3b+2}+ \dfrac{3}{b+3c+2}+ \dfrac{3}{c+3a+2}$
$\rightarrow P\ge 3(\dfrac{1}{a+3b+2}+ \dfrac{1}{b+3c+2}+ \dfrac{1}{c+3a+2})$
$\rightarrow P\ge 3.\dfrac{9}{a+3b+2+b+3c+2+c+3a+2}$
$\rightarrow P\ge 3.\dfrac{9}{4(a+b+c)+6}=3$
Dấu = xảy ra khi $a=b=c=\dfrac{1}{4}$
