Question

cho x+y+z =3

C/m \(\frac{x}{1+y^2}+\frac{y}{1+z^2}+\frac{z}{1+x^2}\)>= 3/2

Answer 1

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

Tương tự: \(\frac{y}{1+z^2}\ge y-\frac{1}{2}yz\) ; \(\frac{z}{1+x^2}\ge z-\frac{1}{2}zx\)

Cộng vế với vế:

\(P\ge x+y+z-\frac{1}{2}\left(xy+yz+zx\right)\ge x+y+z-\frac{1}{6}\left(x+y+z\right)^2=\frac{3}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\)

Answer 2

Đề bài sai.

Phản ví dụ: \(x=-1;y=0;z=4\) thì \(VT=1< \frac{3}{2}\)