Question

cho a, b, c > 0. CMR:

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

Answer 1

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

\(=\dfrac{a^4}{a^2+2ab}+\dfrac{b^4}{b^2+2bc}+\dfrac{c^4}{c^2+2ca}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2}{3}\left(vì\left(a^2+b^2+c^2\ge ab+bc+ca\right)\right)\)

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