\(B=16\left(xy\right)^2+12\left(x^3+y^3\right)+34xy\)
\(B=16\left(xy\right)^2+12\left(x+y\right)^3-36xy\left(x+y\right)+34xy\)
\(B=16\left(xy\right)^2-2xy+12\)
Đặt \(xy=a\Rightarrow0\le a\le\frac{1}{4}\)
\(B=16a^2-2a+12=16\left(a-\frac{1}{16}\right)^2+\frac{191}{16}\ge\frac{191}{16}\)
\(B=16a^2-2a-\frac{1}{2}+\frac{25}{2}=16\left(a-\frac{1}{4}\right)\left(a+\frac{1}{8}\right)+\frac{25}{2}\le\frac{25}{2}\)