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

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

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

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

Do đó \(P\ge15\)

\(\Rightarrow P>0\)

Suy ra P luôn dương

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

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

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

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

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

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

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.

2. Có hai cách nhé

Cách 1: P = xy(x - 2)(y + 6) + 12x² - 24x + 3y² + 18y + 36 
--> P = xy(x - 2)(y + 6) + 12x(x - 2) + 3y(y + 6) + 36 
--> P = [ 12x(x - 2) + 36 ] + xy(x - 2)(y + 6) + 3y(y + 6) 
--> P = 12[x(x - 2) + 3] + y(y + 6).[x(x - 2) + 3] 
--> P = [x(x - 2) + 3].[y(y + 6) + 12] 
--> P = (x² - 2x + 3)(y² + 6y + 12) 
--> P = [(x - 1)² + 2].[(y + 3)² + 3] ≥ 2.3 = 6 > 0 

Dấu " = " xảy ra ⇔ x = 1 ; y = -3 
Vậy MinP = 6 ⇔ x = 1 ; y = -3 

Cách 2: P = xy(x - 2)(y + 6) + 12x² - 24x + 3y² + 18y + 36 
--> P = xy(x - 2)(y + 6) + 12x(x - 2) + 3(y + 3)² + 9 
--> P = x(x - 2)[y(y - 6) + 12] + 3(y + 3)² +9 
--> P = x(x - 2)[(y + 3)² + 3] + 3(y + 3)² + 9 
--> P = x(x - 2)(y + 3)² + 3x(x - 2) + 3(y + 3)² + 9 
--> P = (y + 3)²[x(x - 2) + 3] + 3x(x - 2) + 9 
--> P = (y + 3)²[(x - 1)² + 2] + 3x² - 6x + 9 
--> P = (y + 3)²(x - 1)² + 2(y + 3)² + 3(x - 1)² + 6 ≥ 6 

Dấu " = " xảy ra ⇔ x = 1 ; y = -3 
Vậy MinP = 6 ⇔ x = 1 ; y = -3 

P/S: MinP = 6 > 0 ∀ x, y ∈ R --> P luôn dương ∀ x, y ∈ R 
Mình nghĩ phần CM: "P luôn dương với mọi x,y thuộc R." là hơi thừa :-) 

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Ta có : \(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\)    (*)

\(\Leftrightarrow\frac{x^2}{y^2}+2.\frac{x}{y}.\frac{y}{x}+\frac{y^2}{x^2}-3\left(\frac{x}{y}+\frac{y}{x}\right)+2\ge0\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2-3\left(\frac{x}{y}+\frac{y}{x}\right)+2\ge0\)   (**)

Đặt \(\frac{x}{y}+\frac{y}{x}=t\Rightarrow t\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2\)

Vậy thì \(\left(\frac{x}{y}+\frac{y}{x}\right)^2-3\left(\frac{x}{y}+\frac{y}{x}\right)+2=t^2-3t+2=\left(t-\frac{3}{2}\right)^2-\frac{1}{4}\)

\(\ge\left(2-\frac{3}{2}\right)^2-\frac{1}{4}=0\)

Vậy bất đẳng thức  (**) đúng hay bất đẳng thức (*) đúng

Ta có : 

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

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

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

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

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