Question

Answer 1

\(\dfrac{a^3}{a+2b}+\dfrac{b^3}{b+2c}+\dfrac{c^3}{c+2a}=\dfrac{a^4}{a^2+2ab}+\dfrac{b^4}{b^2+2bc}+\dfrac{c^4}{c^2+2ca}\)

Áp dụng BĐT Svácxơ, ta có:

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

Áp dụng BĐT Cô si, ta có:

\(\left\{{}\begin{matrix}2ab\le a^2+b^2\\2bc\le b^2+c^2\\2ca\le c^2+a^2\end{matrix}\right.\)

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