Question

Answer 1

Lời giải:

$V=\sum (x-\frac{xy^2}{x+y^2})=\sum x - \sum \frac{xy^2}{x+y^2}$

$=3-\sum \frac{xy^2}{x+y^2}$

$\geq 3-\sum \frac{xy^2}{2y\sqrt{x}}$ (theo BĐT AM-GM)

$=3-\frac{1}{2}\sum y\sqrt{x}$

Áp dụng BĐT AM-GM:

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

$\leq \frac{3}{2}+\frac{1}{2}.\frac{1}{3}(x+y+z)^2$

$=\frac{3}{2}+\frac{1}{2}.\frac{1}{3}.3^2=3$

$\Rightarrow V\geq 3 - \frac{3}{2}=\frac{3}{2}$ 

Vậy $V_{\min}=\frac{3}{2}$ khi $x=y=z=1$