Question

[Phùng Gia Bảo]

Cho \(0\le x;y\le1\). Chứng minh: \(\frac{x+y}{2}\le\frac{x}{\sqrt{y+3}}+\frac{y}{\sqrt{x+3}}\le1\)

Answer 1

[Nguyễn Linh Chi]

Với \(0\le x;y\le1\) ta có:

\(\frac{x}{\sqrt{y+3}}+\frac{y}{\sqrt{x+3}}\ge\frac{x}{\sqrt{1+3}}+\frac{y}{\sqrt{1+3}}=\frac{x+y}{2}\)

Dấu "=" xảy ra <=> x = y = 1

Có: \(0\le x;y\le1\)

=> \(0\le x^2\le x\le1;0\le y^2\le y\le1\)

\(\left(\frac{x}{\sqrt{y+3}}+\frac{y}{\sqrt{x+3}}\right)^2\le2\left(\frac{x^2}{y+3}+\frac{y^2}{x+3}\right)\le2\left(\frac{x}{x+y+2}+\frac{y}{x+y+2}\right)\)

\(=2\left(\frac{x+y+2}{x+y+2}-\frac{2}{x+y+2}\right)\le2\left(1-\frac{2}{1+1+2}\right)=1\)

=> \(\sqrt{\frac{x}{\sqrt{y+3}}+\frac{y}{\sqrt{x+3}}}\le1\)

Dấu "=" xảy ra x<=>  = y =1