Question

\(Cho\) \(a,b,c>0\) \(và\) \(abc=1\)

\(CMR:\) \(\dfrac{a}{a^3+a+1}+\dfrac{b}{b^3+b+1}+\dfrac{c}{c^3+c+1}\le1\)

Answer 1

\(cminh:\)\(\dfrac{a}{a^3+a+1}\le\dfrac{4-a}{9}\left(1\right)\)

\(\left(1\right)\Leftrightarrow\)\(9a-\left(4-a\right)\left(a^3+a+1\right)\le0\Leftrightarrow\left(a-1\right)^2\left(a^2-2a-4\right)\le0\left(đúng\right)\Rightarrow\left(1\right)đúng\)

\(tương\) \(tự\Rightarrow\Sigma\dfrac{a}{a^3+a+1}\le\dfrac{4-a+4-b+4-c}{9}=\dfrac{-\left(a+b+c\right)+12}{9}\le\dfrac{-3\sqrt[3]{abc}+12}{9}=\dfrac{-3+12}{9}=1\left(đpcm\right)\)

Answer 2

mik lm đc rồi nha