Đáp án: `P_{min}=\frac{3}{2}⇔x=y=z=1`
Giải thích các bước giải:
Ta có: $3=x+y+z≥3\sqrt[3]{xyz}$
$⇒xyz≤1$
Lại có: $6=(x+1)(y+1)(z+1)≥3\sqrt[3]{(x+1)(y+1)(z+1)}$
$⇒(x+1)(y+1)(z+1)≤8$
Ta có:
`P=\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}`
$≥3\large\sqrt[3]{\frac{1}{x^2+x}.\frac{1}{y^2+y}.\frac{1}{z^2+z}}$
$=3\large\sqrt[3]{\frac{1}{xyz(x+1)(y+1)(z+1)}}$
$≥3\large\sqrt[3]{\frac{1}{8.1}}$
`=3.\frac{1}{2}=\frac{3}{2}`
Dấu bằng xảy ra
$⇔\begin{cases}x=y=z\\x+1=y+1=z+1\\xyz=1\\(x+1)(y+1)(z+1)=8\\\large \frac{1}{x^2+x}=\large \frac{1}{y^2+y}=\large \frac{1}{z^2+z}\\x+y+z=3\end{cases}⇔x=y=z=1$
