ta có :
`(a-b)^2+(b-c)^2+(c-a)^2≥0`
`⇔2(a^2+b^2+c^2)-2(ab+bc+ac)≥0`
`⇔a^2+b^2+c^2≥ab+bc+ac`
`⇔(a+b+c)^2≥3(ab+bc+ac)=3.3=9`
`⇔a+b+c≥3`
mà :
`(a)/(1+2b^3)=a-(2ab^3)/(2b^3+1)≥a-(2ab^3)/(3b^2)=a-(2ab)/3`
tương tự
`⇒(b)/(1+2c^3)≥b-(2bc)/3`
`⇒(c)/(1+2a^3)≥c-(2ac)/3`
`⇒P≥a+b+c-(2(ab+bc+ac))/3`
`⇔P≥3-(2.3)/3`
`⇔P≥1`
`''=''`xẩy ra khi :
`a=b=c=1`