Question

a/(2b+3c) +b/(2c+3a) + c/(2a+3b) >3/5

Answer 1

`a/(2b+3c) +b/(2c+3a) + c/(2a+3b) >=3/5`
Thiếu đk `a,b,c>0`
`a/(2b+3c) +b/(2c+3a) + c/(2a+3b)`
`=a^2/(2ab+3ac)+b^2/(2bc+3ab)+c^2/(2ac+3bc)`
Áp dụng BĐT cosi-schwart: 
`a^2/(2ab+3ac)+b^2/(2bc+3ab)+c^2/(2ac+3bc)>=(a+b+c)^2/(5(ab+bc+ca))=(a^2+b^2+c^2+2ab+2bc+2ca)//(5(ab+bc+ca))`
Áp dụng cosi:`a^2+b^2+c^2>=ab+bc+ca`
`=>a^2/(2ab+3ac)+b^2/(2bc+3ab)+c^2/(2ac+3bc)>=(3(ab+bc+ca))/(5(ab+bc+ca))=3/5`
Dấu "=" xảy ra khi `a=b=c`