`color{#1be01b}{-T}color{#1be01b}{i}color{#1be01b}{t}color{#1be01b}{u}color{#1be01b}{u}color{#1be01b}{u-}color{#1be01c}`

Có: `(a+b+c)[(a)/((b+c)^2)+(b)/((c+a)^2)+(c)/((a+b)^2)]`

`=[(\sqrt{a})^2+(\sqrt{b})^2+(\sqrt{c})^2].[((\sqrt{a})/(b+c))^2+((\sqrt{b})/(c+a))^2+((\sqrt{c})/(a+b))^2] (1)`

Áp dụng BĐT Bunhia có:

`(1)>=(\sqrt{a}.(\sqrt{a})/(b+c)+\sqrt{b}.(\sqrt{b})/(c+a)+\sqrt{c}.(\sqrt{c})/(a+b))^2=((a)/(b+c)+(b)/(c+a)+(c)/(a+b))^2 `

Có: `(a)/(b+c)+(b)/(c+a)+(c)/(a+b)=(a^2)/(a(b+c))+(b^2)/(b(c+a))+(c^2)/(c(a+b)) (2)`

Áp dụng BĐT Cô-si có:

`(2)>=((a+b+c)^2)/(2(ab+bc+ca))>=(3(ab+bc+ca))/(2(ab+bc+ca))=(3)/(2)`

`-> ((a)/(b+c)+(b)/(c+a)+(c)/(a+b))^2>=(9)/(4)`

`-> (a+b+c)[(a)/((b+c)^2)+(b)/((c+a)^2)+(c)/((a+b)^2)]>=(9)/(4)`

`-> (a)/((b+c)^2)+(b)/((c+a)^2)+(c)/((a+b)^2)>=(9)/(4(a+b+c)) (đpcm)`

Dấu `=` xảy ra khi `a=b=c>0`