Xét `A=(abc(5a+16b+27c))/((a+2b)(a+3c)(2b+3c))`
Có: `5a+16b+27c=2a+3a+4b+12b+9c+18c=2(a+2b)+3(a+3c)+6(2b+3c)`
`->A=(abc[2(a+2b)+3(a+3c)+6(2b+3c)])/((a+2b)(a+3c)(2b+3c)) =(2abc)/((a+3c)(2b+3c))+(3abc)/((a+2b)(2b+3c))+(6abc)/((a+2b)(a+3c))`
Theo giả thiết suy ra `a/(abc)+(2b)/(abc)+(3c)/(abc)=6` hay `a+2b+3c=6abc`
Xét mẫu số:
`+) (a+3c)(2b+3c)=2ab+3ac+6bc+9c^2=2ab+3c(a+2b+3c)=2ab+18abc^2=2ab(1+9c^2)`
Tương tự:` (a+2b)(2b+3c)=3ac(1+4b^2); (a+2b)(a+3c)=6bc(1+a^2)`
Nên `A=(2abc)/(2ab(1+9c^2))+(3abc)/(3ac(1+4b^2))+(6abc)/(6bc(1+a^2))`
`=c/(1+9c^2)+b/(1+4b^2)+a/(1+a^2)`

`->ĐPCM`

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Điểm mấu chốt: `5a+16b+27c=2a+3a+4b+12b+9c+18c=2(a+2b)+3(a+3c)+6(2b+3c)`

Cách tìm: dựa vào mẫu số