Question

[Phùng Minh Quân]

Cho \(a>b>c>d>0\) thỏa mãn \(a^2+b^2+c^2=1\)

CMR : \(\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\ge\frac{1}{2}\)

:v

Answer 1

[Đặng Yến Ngọc]

toán lớp 1 gì mà ảo diệu quá...

Answer 2

[Nguyễn Hưng Phát]

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

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

\(=\frac{a^2+b^2+c^2}{2}=\frac{1}{2}\)