By AM-GM: $(\frac{a+3b}{4}*\frac{b+4c}{5}*\frac{c+2a}{3})^{60} \ge a^{55}b^{57}c^{68}$ Remains to prove that $a^{55}b^{57}c^{68} \ge a^{60}b^{60}c^{60}$ After dividing with $a^{55} b^{57} c^{60}$ we get: $c^8 \ge a^5 b^3$ Which is true coz of the condition.

by AM-GM $LHS \ge 60c^{\frac{4}{5}+\frac{1}{3}}b^{\frac{3}{4}+\frac{1}{5}}a^{\frac{1}{4}+\frac{2}{3}}\\ \ge 60cb^{\frac{4}{5}+\frac{1}{3}-1+\frac{3}{4}+\frac{1}{5}}a^{\frac{1}{4}+\frac{2}{3}}\\ \ge 60cba^{\frac{1}{4}+\frac{2}{3}+\frac{4}{5}+\frac{1}{3}-1+\frac{3}{4}+\frac{1}{5}-1}\\ =60cba$

to solve this problem,we can creat a matrix here:let the 2*3 matrix A refers to that the first line contains a b and c,the second line contains 2a,3b and4c.(from the left to the right).And for the 2*3 matrix B,the first line containsa,b and c,the second line contains3b,4c and2a.Now due to c>=b>=a>=0 and S(A)>=S(B),thus the conclusion is correct

And simple substitution $b=a+x$ and $c=a+x+y$ followed by expanding solves it.

See also here

math_science wrote: And simple substitution $b=a+x$ and $c=a+x+y$ followed by expanding solves it.

By AM-GM: \[(a+b+b+b)(b+c+c+c+c)(c+a+a)\geq 4\sqrt[4]{ab^3}5\sqrt[5]{bc^4}3\sqrt[3]{ca^2}=60 a^{\frac14+\frac23}b^{\frac34+\frac15}c^{\frac45+\frac13}\]let's proving: \[60 a^{\frac14+\frac23}b^{\frac34+\frac15}c^{\frac45+\frac13}=60 a^{\frac{11}{12}}b^{\frac{19}{20}}c^{\frac{17}{15}}\geq 60abc\]If and only if: \[\left(c^{\frac{2}{15}}\geq a^{\frac{1}{12}}b^{\frac{1}{20}}\right)^{60}\]\[c^8\geq a^5b^3, \text{but } c^5\geq a^5,c^3\geq b^3, \text{so }c^8\geq a^5b^3\] As desired.

Let $ 0\le a\le b\le c. $ Prove that $$(a+b)(b+2c)(c+3a)\ge 6(2+ \sqrt 3)abc$$$$(a+2b)(b+3c)(c+4a)\ge4(9+4\sqrt 2)abc$$