Question

Cho a,b,c>0. Chứng minh rằng: \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca\)

Answer 1

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

\("="\Leftrightarrow a=b=c\)

Answer 2

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

Dấu "=" xảy ra khi \(a=b=c\)

Answer 3

Cách khác: sử dụng AM-GM:

\(\frac{a^3}{b}+ab\ge2\sqrt{\frac{a^3}{b}\cdot ab}=2a^2\)

TT:\(\frac{b^3}{c}+bc\ge2b^2\);\(\frac{c^3}{a}+ca\ge2c^2\)

Cộng vế theo vế và sử dụng đánh giá quen thuộc \(2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\) ta có đpcm

"="<=>a=b=c