Question

Cho a,b,c là 3 số không âm thỏa mãn \(ab+bc+ca=3\). Chứng minh rằng

\(\dfrac{a}{a^3+b^2+1}+\dfrac{b}{b^3+c^2+1}+\dfrac{c}{c^3+a^2+1}\le1\)

@Akai Haruma

Answer 1

Lightning Farron Nghe nói dạo này anh rảnh rổi lắm phải hông =="

Haha

Answer 2

@Akai Haruma

Answer 3

\(a+b+c\ge ab+bc+ca=3\)

\(\dfrac{a}{a^3+b^2+1}\le\dfrac{a}{a^2+b^2+a}\le\dfrac{a}{2ab+a}=\dfrac{1}{2b+1}\)

\(\sum\dfrac{1}{2a+1}\le1\Leftrightarrow3-\sum\dfrac{2a}{2a+1}\le1\)

\(\sum\dfrac{a}{2a+1}\ge1\)....

Answer 4

Bác Truy kích làm rồi mà. Nhờ a làm gì nữa.

Answer 5

Since
it's enough to prove that
or
or

or
and since
and by Schur
it's enough to prove that
or
which is obvious.