Question

[Nguyễn Băng Băng]

Cho x y la số thực không âm Chứng minh

\(\sqrt{\frac{x}{y+1}}+\sqrt{\frac{y}{x+1}}+\sqrt{\frac{1}{x+y}}\ge2\)

Answer 1

[kudo shinichi]

Ta có: \(x+\left(y+1\right)\ge2.\sqrt{x.\left(y+1\right)}=2.\sqrt{xy+x}\)

\(y+\left(x+1\right)\ge2.\sqrt{y.\left(x+1\right)}=2.\sqrt{xy+y}\)

\(1+\left(x+y\right)\ge2.\sqrt{x+y}\)

Ta có: \(\sqrt{\frac{x}{y+1}}+\sqrt{\frac{y}{x+1}}+\sqrt{\frac{1}{x+y}}\)

\(=\frac{\sqrt{x}}{\sqrt{y+1}}+\frac{\sqrt{y}}{\sqrt{x+1}}+\frac{1}{\sqrt{x+y}}\)

\(=\frac{x}{\sqrt{yx+x}}+\frac{y}{\sqrt{xy+y}}+\frac{1}{\sqrt{x+y}}\)

\(=\frac{2x}{2\sqrt{yx+x}}+\frac{2y}{2\sqrt{xy+y}}+\frac{2}{2\sqrt{x+y}}\)

\(\ge\frac{2x}{x+y+1}+\frac{2y}{x+y+1}+\frac{21}{x+y+1}=\frac{2\left(x+y+1\right)}{x+y+1}=2\)

                                                                                        đpcm

Tham khảo nhé~