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H(x)=x2+y2−xy+x+y+1

\(\Rightarrow 12 H \left(\right. x \left.\right) = 12 \left(\right. x^{2} + y^{2} - x y - x + y + 1 \left.\right)\)

\(\Rightarrow 12 H \left(\right. x \left.\right) = 12 x^{2} + 12 y^{2} - 12 x y - 12 x + 12 y + 12\)

\(\Rightarrow 12 H \left(\right. x \left.\right) = \left(\right. 12 x^{2} - 12 x y + 3 y^{2} - 12 x + 6 y + 3 \left.\right) + \left(\right. 9 y^{2} + 6 y + 9 \left.\right)\)

\(\Rightarrow 12 H \left(\right. x \left.\right) = 3 \left(\right. 4 x^{2} - 4 x y + y^{2} - 4 x + 2 y + 1 \left.\right) + \left(\right. 9 y^{2} + 6 y + 1 \left.\right) + 8\)

\(\Rightarrow 12 H \left(\right. x \left.\right) = 3 \left[\right. \left(\left(\right. 2 x \left.\right)\right)^{2} + y^{2} + 1^{2} - 2 \cdot 2 x \cdot y - 2 \cdot 2 x \cdot 1 + 2 \cdot y \cdot 1 \left]\right. + \left[\right. \left(\left(\right. 3 y \left.\right)\right)^{2} + 2 \cdot 3 y \cdot 1 + 1^{2} \left]\right. + 8\)

\(\Rightarrow 12 H \left(\right. x \left.\right) = 3 \left(\left(\right. 2 x - y - 1 \left.\right)\right)^{2} + \left(\left(\right. 3 y + 1 \left.\right)\right)^{2} + 8\)

\(\Rightarrow H \left(\right. x \left.\right) = \frac{3 \left(\left(\right. 2 x - y - 1 \left.\right)\right)^{2} + \left(\left(\right. 3 y + 1 \left.\right)\right)^{2} + 8}{12} = \frac{\left(\left(\right. 2 x - y - 1 \left.\right)\right)^{2}}{4} + \frac{\left(\left(\right. 3 y + 1 \left.\right)\right)^{2}}{12} + \frac{2}{3}\)

Ta có: \(\left{\right. \frac{\left(\left(\right. 2 x - y - 1 \left.\right)\right)^{2}}{4} \geq 0 \forall x , y \\ \frac{\left(\left(\right. 3 y + 1 \left.\right)\right)^{2}}{12} \geq 0 \forall y\)

\(\Rightarrow H \left(\right. x \left.\right) = \frac{\left(\left(\right. 2 x - y - 1 \left.\right)\right)^{2}}{4} + \frac{\left(\left(\right. 3 y + 1 \left.\right)\right)^{2}}{12} + \frac{2}{3} \geq \frac{2}{3} \forall x , y\)

Dấu "=" xảy ra:

\(\left{\right. 2 x - y - 1 = 0 \\ 3 y + 1 = 0\)

\(\Leftrightarrow \left{\right. 2 x + \frac{1}{3} - 1 = 0 \\ y = - \frac{1}{3}\)

\(\Leftrightarrow \left{\right. x = \frac{1}{3} \\ y = - \frac{1}{3}\)

Vậy: ... 

H(x)=x2+y2−xy+x+y+1

\(\Rightarrow 12 H \left(\right. x \left.\right) = 12 \left(\right. x^{2} + y^{2} - x y - x + y + 1 \left.\right)\)

\(\Rightarrow 12 H \left(\right. x \left.\right) = 12 x^{2} + 12 y^{2} - 12 x y - 12 x + 12 y + 12\)

\(\Rightarrow 12 H \left(\right. x \left.\right) = \left(\right. 12 x^{2} - 12 x y + 3 y^{2} - 12 x + 6 y + 3 \left.\right) + \left(\right. 9 y^{2} + 6 y + 9 \left.\right)\)

\(\Rightarrow 12 H \left(\right. x \left.\right) = 3 \left(\right. 4 x^{2} - 4 x y + y^{2} - 4 x + 2 y + 1 \left.\right) + \left(\right. 9 y^{2} + 6 y + 1 \left.\right) + 8\)

\(\Rightarrow 12 H \left(\right. x \left.\right) = 3 \left[\right. \left(\left(\right. 2 x \left.\right)\right)^{2} + y^{2} + 1^{2} - 2 \cdot 2 x \cdot y - 2 \cdot 2 x \cdot 1 + 2 \cdot y \cdot 1 \left]\right. + \left[\right. \left(\left(\right. 3 y \left.\right)\right)^{2} + 2 \cdot 3 y \cdot 1 + 1^{2} \left]\right. + 8\)

\(\Rightarrow 12 H \left(\right. x \left.\right) = 3 \left(\left(\right. 2 x - y - 1 \left.\right)\right)^{2} + \left(\left(\right. 3 y + 1 \left.\right)\right)^{2} + 8\)

\(\Rightarrow H \left(\right. x \left.\right) = \frac{3 \left(\left(\right. 2 x - y - 1 \left.\right)\right)^{2} + \left(\left(\right. 3 y + 1 \left.\right)\right)^{2} + 8}{12} = \frac{\left(\left(\right. 2 x - y - 1 \left.\right)\right)^{2}}{4} + \frac{\left(\left(\right. 3 y + 1 \left.\right)\right)^{2}}{12} + \frac{2}{3}\)

Ta có: \(\left{\right. \frac{\left(\left(\right. 2 x - y - 1 \left.\right)\right)^{2}}{4} \geq 0 \forall x , y \\ \frac{\left(\left(\right. 3 y + 1 \left.\right)\right)^{2}}{12} \geq 0 \forall y\)

\(\Rightarrow H \left(\right. x \left.\right) = \frac{\left(\left(\right. 2 x - y - 1 \left.\right)\right)^{2}}{4} + \frac{\left(\left(\right. 3 y + 1 \left.\right)\right)^{2}}{12} + \frac{2}{3} \geq \frac{2}{3} \forall x , y\)

Dấu "=" xảy ra:

\(\left{\right. 2 x - y - 1 = 0 \\ 3 y + 1 = 0\)

\(\Leftrightarrow \left{\right. 2 x + \frac{1}{3} - 1 = 0 \\ y = - \frac{1}{3}\)

\(\Leftrightarrow \left{\right. x = \frac{1}{3} \\ y = - \frac{1}{3}\)

Vậy: ... 

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