Ta có : 

\(B=x\left(x-2\right)y\left(y+6\right)+12x^2-24x+3y^2+18y+36\)

\(=\left(x^2-2x\right)\left(y^2+6y\right)+12\left(x^2-2x\right)+3\left(y^2+6y+12\right)+12\)

\(=\left(x^2-2x\right)\left(y^2+6y+12\right)+3\left(y^2+6y+12\right)+12\)

\(=\left(x^2-2x+3\right)\left(y^2+6y+12\right)+12\)

\(=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]+12\ge2.3+12=18\)

\(xy\left(x-2\right)\left(y+6\right)+12x^2-24x+3y^2+18y+2045.\)

\(=\left(x^2-2x\right)\left(y^2+6y\right)+12\left(x^2-2x\right)+3\left(y^2+6y\right)+2045\)

\(=\left[\left(x^2-2x\right)\left(y^2+6y\right)+3\left(y^2+6y\right)\right]+12\left(x^2-2x+3\right)+2009.\)

\(=\left(x^2-2x+3\right)\left(y^2+6x\right)+12\left(x^2-2x+3\right)+2009\)

\(=\left(x^2-2x+3\right)\left(y^2+6x+12\right)+2009\)

\(=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]+2009\)

Ta có: \(\left(x-1\right)^2\ge0\forall x\Leftrightarrow\left(x-1\right)^2+2\ge2\)

\(\left(y+3\right)^2\ge0\forall y\Leftrightarrow\left(y+3\right)^2+3\ge3\)

Suy ra \(B=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]+2009\ge2.3+2009=2015\)

Vậy GTNN của B=2015 khi x=1, y=-3.

\(P=xy\left(x-2\right)\left(y+6\right)+12x^2-24x+3y^2+18y+36\)

\(=\left(x^2-2x\right)\left(y^2+6y\right)+\left(12x^2+24x+12\right)+\left(3y^2+18y+9\right)+15\)

\(=\left[\left(x-1\right)^2-1\right]\left[\left(y+3\right)^2-9\right]+12\left(x-1\right)^2+3\left(y+3\right)^2+15\)

\(=3\left(x-1\right)^2+2\left(y+3\right)^2+15\)

Do đó \(P\ge15\)

\(\Rightarrow P>0\)

Suy ra P luôn dương

1/ Với số dương ta luôn có \(\frac{x}{y}+\frac{y}{x}\ge2\) (Cauchy hoặc quy đồng chuyển vế sẽ chứng minh được dễ dàng). Ta cần chứng minh:

\(\frac{x^2}{y^2}+\frac{y^2}{x^2}+2.\frac{x}{y}.\frac{y}{x}+2\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2+2\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\) (1)

Đặt \(\frac{x}{y}+\frac{y}{x}=a\ge2\) thì (1) trở thành:

\(a^2+2\ge3a\Leftrightarrow a^2-3a+2\ge0\Leftrightarrow\left(a-1\right)\left(a-2\right)\ge0\) (2)

Do \(a\ge2\Rightarrow\left\{{}\begin{matrix}a-1>0\\a-2\ge0\end{matrix}\right.\Rightarrow\left(a-1\right)\left(a-2\right)\ge0\)

\(\Rightarrow\left(2\right)\) đúng, vậy BĐT được chứng minh. Dấu "=" xảy ra khi \(x=y\)

2/ \(B=\left(x^2-2x\right)\left(y^2+6y\right)+12\left(x^2-2x\right)+3\left(y^2+6y\right)+2045\)

\(B=\left(x^2-2x\right)\left(y^2+6y+12\right)+3\left(y^2-6y+12\right)-36+2045\)

\(B=\left(x^2-2x+3\right)\left(y^2+6y+12\right)+2009\)

\(B=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]+2009\)

Do \(\left\{{}\begin{matrix}\left(x-1\right)^2+2\ge2\\\left(y+3\right)^2+3\ge3\end{matrix}\right.\)

\(\Rightarrow B\ge2.3+2009=2015\)

\(\Rightarrow B_{min}=2015\) khi \(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)