Question

Cho a,b,c>0 tm a+b+c=6

CMR \(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\ge3\)

Answer 1

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

\(=\dfrac{a^3+ab^2-ab^2}{a^2+b^2}+\dfrac{b^3+bc^2-bc^2}{b^2+c^2}+\dfrac{c^3+ca^2-ca^2}{c^2+a^2}\)

\(=\dfrac{a\left(a^2+b^2\right)}{a^2+b^2}+\dfrac{b\left(b^2+c^2\right)}{b^2+c^2}+\dfrac{c\left(c^2+a^2\right)}{c^2+a^2}-\left(\dfrac{ab^2}{a^2+b^2}+\dfrac{bc^2}{b^2+c^2}+\dfrac{ca^2}{c^2+a^2}\right)\)\(=a+b+c-\left(\dfrac{ab^2}{a^2+b^2}+\dfrac{bc^2}{b^2+c^2}+\dfrac{ca^2}{c^2+a^2}\right)\)\(\ge6-\left(\dfrac{ab^2}{2ab}+\dfrac{bc^2}{2bc}+\dfrac{ca^2}{2ca}\right)=6-\dfrac{a+b+c}{2}=6-\dfrac{6}{2}=3\)      (dpcm)Dau = xay ra khi a=b=c=2