\(P=\left(4x^2y^2\right)\left(-3x^6y^2\right)=-12x^8y^4\)
\(Q=\left(\dfrac{1}{2}x^5y\right)\left(-2xy\right)^3=\left(\dfrac{1}{2}x^5y\right)\left(-8x^3y^3\right)=-4x^8y^4\)
\(P+Q+25=-16x^8y^4+25\le25\)
Dấu ''='' xảy ra khi \(\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)