Question

Cho a,b,c > 0. CMR: \(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\text{≥}ab+bc+ca\)

Answer 1

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

Answer 2

BĐT:\(a,b,c>0\Rightarrow\left(ab+bc+ac\right)\ne0\)

\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}\)

\(\ge\dfrac{\left(ab+bc+ac\right)^2}{ab+bc+ac}=ab+bc+ac\)