Question

Answer 1

Áp dụng BĐT Cô-si:

\(\dfrac{a^3}{b}+ab\ge2\sqrt{\dfrac{a^4b}{b}}=2a^2\)

Tương tự: \(\dfrac{b^3}{c}+bc\ge2b^2\); \(\dfrac{c^3}{a}+ca\ge2c^2\)

Cộng vế:

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ca\ge2\left(a^2+b^2+c^2\right)\) (1)

Mặt khác:

\(a^2+b^2\ge2ab\) ; \(b^2+c^2\ge2bc\); \(c^2+a^2\ge2ca\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\) (2)

(1);(2) \(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)

Dấu "=" xảy ra khi \(a=b=c\)