Chủ đề này có 2 trả lời

[VuQuyDat]

x,y,z>0;  x+y+z=3

Tìm min            

                             $\frac{x+1}{y^2+1}+\frac{y+1}{z^2+1}+\frac{z+1}{x^2+1}$

[Khoa Linh]

Ta có: $\frac{x+1}{y^2+1}=(x+1)-\frac{y^2(x+1)}{y^2+1}\geq x+1-\frac{y(x+1)}{2}$

Suy ra $\sum \frac{x+1}{y^2+1}\geq \frac{x+y+z-(xy+yz+zx)}{2}+3\geq 3$

[Kiratran]

$x+1+y+1+z+1-\frac{y^2(x+1)}{y^2+1}-\frac{z^2(y+1)}{z^2+1}-\frac{x^2(z+1)}{x^2+1}\geq 6-\frac{y(x+1)}{2}-\frac{z(y+1)}{2}-\frac{x(z+1)}{2}\geq 6-3=3$