Question

CMR với a,b,c >0 \(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\ge\dfrac{a+b+c}{2}\)

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

Cộng (1), (2), (3) ta có:

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

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)