`P=x^3+y^3+2x^2y^2`
`P=(x+y)^3-3xy(x+y)+2x^2y^2`
`P=1-3xy+2x^2y^2`
`P=2(x^2y^2-3/2xy+1/2)`
`P=2[(xy)^2-2. xy. 3/4+9/16-1/16]`
`P=2(xy-3/4)^2-1/8`
`P=2[x(1-x)-3/4]^2-1/8`
`P=2(x-x^2-3/4)^2-1/8`
`P=2(x^2-x+3/4)^2-1/8`
`P=2(x^2-2.x. 1/2+1/4+1/2)^2-1/8`
`P=2[(x-1/2)^2+1/2]^2-1/8`
Do `(x-1/2)^2>=0`
`=> (x-1/2)^2+1/2>=1/2`
`=> 2[(x-1/2)^2+1/2]^2>=2.1/4=1/2`
`=> 2[(x-1/2)^2+1/2]^2-1/8>=3/8`
Dấu = xảy ra khi $\begin{cases}x+y=1\\x=\dfrac{1}{2}\end{cases} ⇔ x=y=\dfrac{1}{2}$
Vậy `P_(min)=3/8<=>x=y=1/2`
