`a/(b+c)+b/(c+a)+c/(a+b)`
`=(a^2)/(a(b+c))+(b^2)/(b(c+a))+(c^2)/(c(a+b))`
`=(a^2)/(ab+ac)+b^2/(bc+ab)+c^2/(ac+bc)`
Áp dụng bđt Svac-xơ ta có:
`(a^2)/(ab+ac)+b^2/(bc+ab)+c^2/(ac+bc)`
`>=((a+b+c)^2)/(ab+ac+bc+ab+ac+bc)`
`=(a+b+c)^2/(2(ab+bc+ca))`
Mặt khác `(a-b)^2+(b-c)^2+(c-a)^2>=0` với `∀a;b;c`
`<=> 2(a^2+b^2+c^2)>=2(ab+bc+ca)`
`<=> a^2+b^2+c^2>=ab+bc+ca`
`<=> a^2+b^2+c^2+2ab+2bc+2ca>=3ab+3bc+3ca`
`<=> (a+b+c)^2>=3(ab+bc+ca)`
`<=> ab+bc+ca<=(a+b+c)^2/3`
`=> (a^2)/(ab+ac)+b^2/(bc+ab)+c^2/(ac+bc)`
`>= (a+b+c)^2/(2. 1/3 (a+b+c)^2)`
`= 3/2`
Dấu = xảy ra khi `a=b=c`
