Đáp án:
Giải thích các bước giải:
$P= \frac{(1+2x)(1+2y)+(1+2y)(1+2z)+(1+2x)(1+2x)}{(1+2x)(1+2y)(1+2z)}=\frac{4(xy+yz+zx)+4(x+y+z)+3}{4(xy+yz+zx)+2(x+y+z)+8xyz+1}$
$P=\frac{4(xy+yz+zx)+2(x+y+z)+2(x+y+z)+3}{4(xy+yz+zx)+2(x+y+z)+9} \geq \frac{4(xy+yz+zx)+2(x+y+z)+2.3\sqrt[3]{xyz}+3}{4(xy+yz+zx)+2(x+y+z)+9} =\frac{4(xy+yz+zx)+2(x+y+z)+9}{4(xy+yz+zx)+2(x+y+z)+9}=1$
$P_{min}=1$ khi $x=y=z=1$