Cho a,b,c là 3 số không âm thỏa mãn \(ab+bc+ca=3\). Chứng minh rằng
\(\dfrac{a}{a^3+b^2+1}+\dfrac{b}{b^3+c^2+1}+\dfrac{c}{c^3+a^2+1}\le1\)
\(a+b+c\ge ab+bc+ca=3\)
\(\dfrac{a}{a^3+b^2+1}\le\dfrac{a}{a^2+b^2+a}\le\dfrac{a}{2ab+a}=\dfrac{1}{2b+1}\)
\(\sum\dfrac{1}{2a+1}\le1\Leftrightarrow3-\sum\dfrac{2a}{2a+1}\le1\)
\(\sum\dfrac{a}{2a+1}\ge1\)....
Bác Truy kích làm rồi mà. Nhờ a làm gì nữa.
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