ta có :
`x^2/(y+1)+(y+1)/4≥2\sqrt(x^2/(y+1).(y+1)/4)=2x/2=x`
tương tự
`⇒y^2/(z+1)+(z+1)/4≥y`
`⇒z^2/(x+1)+(x+1)/4≥z`
`⇒F+(x+z+y+1+1+1)/4≥x+y+z`
`⇒F≥(x+y+z)-(x+z+y)/4-3/4`
`⇒F≥3/4 (x+y+z-1)`
`⇒F≥3/4 (3∛xyz-1)`
`⇒F≥3/4 (3-1)=3/2`
`''=''`xẩy ra khi :
`x=y=z=1`