Question

Cho a,b,c > 0 . CM :

\(\dfrac{1}{a^3+b^3+abc}+\dfrac{1}{b^3+c^3+abc}+\dfrac{1}{c^3+a^3+abc}\le\dfrac{1}{abc}\)

Answer 1

\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\text{≥}\) \(\left(a+b\right)ab\)

⇒ \(a^3+b^3+abc\text{≥}\left(a+b\right)ab+abc=ab\left(a+b+c\right)\)

Tương tự : \(b^3+c^3+abc\text{ ≥}\left(b+c\right)bc+abc=bc\left(a+b+c\right)\)

\(c^3+a^3+abc\text{ ≥}\left(a+c\right)ac+abc=ac\left(a+b+c\right)\)

⇒ \(VT\text{ }\text{≤}\dfrac{1}{a+b+c}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=\dfrac{1}{a+b+c}.\dfrac{a+b+c}{abc}=\dfrac{1}{abc}\)