Áp dụng BĐT `Cauchy-schwars`
`P=a^3/\sqrt(b^2+3)+b^3/\sqrt(c^2+3)+c^3/\sqrt(a^2+3)`
`P=a^4/[a\sqrt(b^2+3)]+b^4/[b\sqrt(c^2+3)]+c^4/[c\sqrt(a^2+3)]`
`P\ge (a^2+b^2+c^2)^2/[a\sqrt(b^2+3)+b\sqrt(c^2+3)+c\sqrt(a^2+3)]`
`P\ge (a^2+b^2+c^2)^2/\sqrt[(a^2+b^2+c^2)(b^2+3+c^2+3+a^2+3)]`
`P\ge (a^2+b^2+c^2)^2/[2(a^2+b^2+c^2)]=3/2`
Dấu `=` xảy ra `⇔a=b=c=1`
Vậy $Min_P=\dfrac{3}{2}⇔a=b=c=1$