Question

[EDOGAWA CONAN]

cho các số thực dương x , y thỏa mãn : x + y = 2 . CMR \(\frac{x}{1+y^2}+\frac{y}{1+x^2}\ge1\)

Answer 1

[Mashiro Shiina]

@Nguyễn Việt Lâm e xin góp 1 cách khác

\(\frac{x}{1+y^2}+\frac{y}{1+x^2}\ge\frac{x^2}{x+xy^2}+\frac{y^2}{y+yx^2}\ge\frac{\left(x+y\right)^2}{x+y+xy\left(x+y\right)}=\frac{\left(x+y\right)^2}{x+y+2xy}\ge\frac{\left(x+y\right)^2}{x+y+\frac{\left(x+y\right)^2}{2}}=1\)\("="\Leftrightarrow x=y=1\)

Answer 2

[Nguyễn Việt Lâm]

\(VT=\frac{x}{1+y^2}+\frac{y}{1+x^2}=x-\frac{xy^2}{1+y^2}+y-\frac{x^2y}{1+x^2}\)

\(VT\ge x+y-\frac{xy^2}{2y}-\frac{x^2y}{2x}=x+y-xy\ge x+y-\frac{\left(x+y\right)^2}{4}=1\) (đpcm)

Dấu "=" xảy ra khi \(x=y=1\)