Question

Cho x,y > 0 và xy = 1. CMR: \(B\ge1\) với \(B=\frac{x^3}{y+1}+\frac{y^3}{x+1}\)

Answer 1

\(\frac{x^3}{y+1}+\frac{y+1}{4}+\frac{1}{2}\ge3\sqrt[3]{\frac{x^3\left(y+1\right)}{8\left(y+1\right)}}=\frac{3}{2}x\)

Tương tự: \(\frac{y^3}{x+1}+\frac{x+1}{4}+\frac{1}{2}\ge\frac{3}{2}y\)

Cộng vế với vế:

\(B+\frac{x+y+2}{4}+1\ge\frac{3}{2}\left(x+y\right)\)

\(\Rightarrow B\ge\frac{5}{4}\left(x+y\right)-\frac{3}{2}\ge\frac{5}{4}.2\sqrt{xy}-\frac{3}{2}=1\)

Dấu "=" xảy ra khi \(x=y=1\)