Question

Cho a, b, c > 0 . CMR :

A= \(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\ge\dfrac{\left(a+b+c\right)}{2}\)

Answer 1

- Ta có: \(\dfrac{a^3}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2}\ge a-\dfrac{ab^2}{2ab}=a-\dfrac{b}{2}\left(1\right)\)

- Tương tự: \(\dfrac{b^3}{b^2+c^2}\ge b-\dfrac{c}{2}\left(2\right)\); \(\dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\left(3\right)\)

- Lấy \(\left(1\right)+\left(2\right)+\left(3\right)\), ta được:

\(A\ge a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\left(đpcm\right)\)

- Dấu "=" xảy ra khi \(a=b=c\)