Ta có \(x^4+y^4-1=xy\left(3-2xy\right)\)
\(\Leftrightarrow x^4+y^4-1=3xy-2x^2y^2\)
\(\Leftrightarrow x^4+2x^2y^2+y^4=3xy+1\)
\(\Leftrightarrow\left(x^2+y^2\right)^2=3xy+1\)
Vì \(\left(x^2+y^2\right)^2\ge0\forall x;y\)
\(\Rightarrow3xy+1\ge0\)
\(\Leftrightarrow xy\ge-\frac{1}{3}\)
\(\Leftrightarrow P\ge-\frac{1}{3}\)
Dấu "=" tại x = y = 0