Theo $AM-GM$ ta có:
$a+3b=a+b+b+b\geq 4\sqrt[4]{ab^{3}}$
$b+4c=b+c+c+c+c\geq 5\sqrt[5]{bc^{4}}$
$c+2a=c+a+a\geq 3\sqrt[3]{ca^{2}}$
Nhân vế theo vế: $(a+3b)(b+4c)(c+2a)\geq 60\sqrt[4]{ab^{3}}\sqrt[5]{bc^{4}}\sqrt[3]{ca^{2}}$
$=60a^{\frac{11}{12}}b^{\frac{19}{20}}c^{\frac{17}{15}}$$=60abca^{\frac{-1}{12}}b^{\frac{-1}{20}}c^{\frac{2}{15}}$$\geq 60abcc^{\frac{-1}{12}}c^{\frac{-1}{20}}c^{\frac{2}{15}}$$\geq 60abc$
Cho mk CTLHN nhé!