Question

Answer 1

Lời giải:

Áp dụng BĐT AM-GM:

$B=\sum \frac{1}{(x^2+1)+(x^2+y^2)+2}\leq \sum \frac{1}{2x+2xy+2}$

$=\frac{1}{2}\sum \frac{1}{xy+x+1}$

$=\frac{1}{2}(\frac{1}{xy+x+1}+\frac{1}{yz+y+1}+\frac{1}{zx+z+1})$

$=\frac{1}{2}(\frac{z}{xyz+xz+z}+\frac{xz}{yz.xz+y.xz+xz}+\frac{1}{xz+z+1})$

$=\frac{1}{2}(\frac{z}{1+xz+z}+\frac{xz}{z+1+xz}+\frac{1}{xz+z+1})$

$=\frac{1}{2}.\frac{z+xz+1}{1+xz+z}=\frac{1}{2}$

Vậy $B_{\max}=\frac{1}{2}$. Giá trị này đạt tại $x=y=z=1$