Question

Answer 1

Ủa có gợi ý rồi thì làm dễ mà ;))

\(a,b,c>0;a+b+c=3\)

Ta có: \(\sum\dfrac{a}{a^3+b+c}=\sum\dfrac{a\left(\dfrac{1}{a}+b+c\right)}{\left(a^3+b+c\right)\left(\dfrac{1}{a}+b+c\right)}\le^{CBS}\sum\dfrac{a\left(\dfrac{1}{a}+b+c\right)}{\left(a+b+c\right)^2}=\dfrac{1}{9}\sum\left(1+ab+ac\right)\)

\(=\dfrac{1}{3}+\dfrac{2}{9}\left(ab+bc+ca\right)\le\dfrac{1}{3}+\dfrac{2}{9}.\dfrac{\left(a+b+c\right)^2}{3}=1\left(đpcm\right)\)

Đẳng thức xảy ra khi \(a=b=c=1\)