Question

Với a>0,b>0,c>0
Cmr: \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ac\)

Gợi ý:  áp dụng bđt \(^{a^3+b^3\ge ab\left(a+b\right)}\)

Answer 1

áp dụng BĐT : \(x^3+y^3\ge xy\left(x+y\right)\) ta có:

\(a^3+b^3\ge ab\left(a+b\right)\)

\(\Leftrightarrow\frac{a^3}{b}+b^2\ge a\left(a+b\right)\)  (vì b>0)

\(\Leftrightarrow\frac{a^3}{b}+b^2\ge a^2+ab\)     (1)

c/m tương tự ta đc: \(\frac{b^3}{c}+c^2\ge b^2+bc\)  (2)

\(\frac{c^3}{a}+a^2\ge c^2+ca\)    (3)

Từ (1),(2),(3)=> \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca\) =>đpcm

Answer 2

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

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