Ta có :

\(B=x^2+xy+y^2-2x-3y+2019\)

\(\Leftrightarrow4B=4x^2+4xy+4y^2-8x-12y+8076\)

\(\Leftrightarrow4B=\left(4x^2+4xy+y^2\right)-4\left(2x+y\right)+4+3y^2-4y+4022\)

\(\Leftrightarrow2B=\left(2x+y\right)^2-4\left(2x+y\right)+4+3\left(y^2-\frac{4}{3}y+\frac{4}{9}\right)+\frac{12062}{3}\)

\(\Leftrightarrow2B=\left(2x+y-2\right)^2+3\left(y-\frac{2}{3}\right)^2+\frac{12062}{3}\ge\frac{12062}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{2}{3}\)

Bạn kiểm tra lại nhé, mình k chắc có đúng k nữa !

C/m: \(\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)

\(\Rightarrow2x^2+xy+2y^2\ge\dfrac{5}{4}\left(x^2+2xy+y^2\right)\)

\(\Leftrightarrow8x^2+4xy+8y^2\ge5x^2+10xy+5y^2\)

\(\Leftrightarrow3\left(x-y\right)^2\ge0\left(LĐ\right)\)

Vậy \(\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)

CMTT: \(\sqrt{2y^2+yz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)\);

\(\sqrt{2z^2+zx+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)

Vậy H=\(\sqrt{2x^2+xy+2y^2}+\sqrt{2y^2+yz+2z^2}+\sqrt{2z^2+xz+2z^2}\ge\sqrt{5}\left(x+y+z\right)=2019\)H_min=2019\(\Leftrightarrow x=y=z=\dfrac{\dfrac{2019}{\sqrt{5}}}{3}\)

Khos quas

\(\sqrt{2x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\frac{2x+y+z}{2}\)

cmtt => GTLN

Tìm max:

Ta có:

\(\sqrt{2x+yz}=\sqrt{x\left(x+y+z\right)+xz}=\sqrt{\left(x+y\right)\left(x+z\right)}\)

\(\le\frac{2x+y+z}{2}\left(1\right)\)

Tương tự ta có: \(\hept{\begin{cases}\sqrt{2y+zx}\le\frac{2y+z+x}{2}\left(2\right)\\\sqrt{2z+xy}\le\frac{2z+x+y}{2}\left(3\right)\end{cases}}\)

Cộng (1), (2), (3) vế theo vế ta được

\(A\le\frac{2x+y+z}{2}+\frac{2y+z+x}{2}+\frac{2z+x+y}{2}=2\left(x+y+z\right)=4\)

Dấu = xảy ra khi \(x=y=z=\frac{2}{3}\)

Tìm min:

Ta có: \(\hept{\begin{cases}\sqrt{2x+yz}\ge0\\\sqrt{2y+zx}\ge0\\\sqrt{2z+xy}\ge0\end{cases}}\)

\(\Rightarrow A\ge0\)

Dấu = xảy ra khi \(\left(x,y,z\right)=\left(-2,2,2;2,-2,2;2,2,-2\right)\)

\(A=x^2+2y^2-2xy-2y-2x+2019\)

\(A=x^2+y^2+y^2-2xy+2y-4y-2x+2019\)

\(A=\left(x^2-2xy+y^2\right)-\left(2x-2y\right)+1+y^2-4y+4+2014\)

\(A=\left(x-y\right)^2-2\left(x-y\right)+1+\left(y-2\right)^2+2014\)

\(A=\left(x-y-1\right)^2+\left(y-2\right)^2+2014\ge2014\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2-1=0\\y=2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=2\end{cases}}}\)

\(4B=4x^2+4xy+4y^2-8x-12y+8076\)

= \(\left(2y\right)^2-4y\left(3-x\right)+\left(3-x\right)^2-\left(3-x\right)^2\)

\(+\left(2x\right)^2-8x+8076\)

= \(\left(2y-3+x\right)^2+3x^2-2x+8076\)

đến đây thì dễ rồi

\(A=x^2+y^2+1-2xy-2x+2y+y^2-4y+4+2014\)

\(=\left(x-y-1\right)^2+\left(y-2\right)^2+2014\ge2014\)

\(\Rightarrow A_{min}=2014\) khi \(\left\{{}\begin{matrix}y-2=0\\x-y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=2\\x=3\end{matrix}\right.\)