Cho `a+b+c=3` va `a,b,c>0` CM: `a^3 /(b(2c+a)) +b^3 /(c(2a+b)) +c^3 /(a(2b+c)) >= 1`

Áp dụng BĐT Cauchy ta có:

`(9a^3)/(b(2c+a))+3b+(2c+a)>=3\root{3}{(9a^3)/(b(2c+a)).3b.(2c+a)}>=3\root{3}{27a^3}=9a`

`(9b^3)/(c(2a+b))+3c+(2a+b)>=3\root{3}{(9b^3)/(c(2a+b)).3c.(2a+b)}>=3\root{3}{27b^3}=9b`

`(9c^3)/(a(2b+c))+3a+(2b+c)>=3\root{3}{(9c^3)/(a(2b+c)).3a.(2b+c)}>=3\root{3}{27c^3}=9c`

Cộng vế theo vế ta được:

`(9a^3)/(b(2c+a))+3b+(2c+a)+(9b^3)/(c(2a+b))+3c+(2a+b)+(9c^3)/(a(2b+c))+3a+(2b+c)>=9a+9b+9c`

`<=>9[(a^3)/(b(2c+a))+(b^3)/(c(2a+b))+(c^3)/(a(2b+c))]+6a+6b+6c>=9a+9b+9c`

`<=>9[(a^3)/(b(2c+a))+(b^3)/(c(2a+b))+(c^3)/(a(2b+c))]>=3a+3b+3c`

`<=>9[(a^3)/(b(2c+a))+(b^3)/(c(2a+b))+(c^3)/(a(2b+c))]>=3(a+b+c)`

Theo đề ta có `a+b+c=3`

`=>9[(a^3)/(b(2c+a))+(b^3)/(c(2a+b))+(c^3)/(a(2b+c))]>=9`

`<=>(a^3)/(b(2c+a))+(b^3)/(c(2a+b))+(c^3)/(a(2b+c))>=1(đpcm)`

Dấu "`=`" xảy ra`<=>a=b=c=1`