\(x^2+2y^2+z^2-2\left(xy+2y+2z+8\right)=0\)

\(pt\Leftrightarrow x^2+2y^2+z^2-2xy+4y+4z+16=0\)

\(\Leftrightarrow x^2-2xy+y^2+y^2+4y+4+z^2+4z+4+8=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y+2\right)^2+\left(z+2\right)^2+8=0\)

Dễ thấy: \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y+2\right)^2\ge0\\\left(z+2\right)^2\ge0\end{cases}}\)

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2+\left(z+2\right)^2\ge0\)

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2+\left(z+2\right)^2+8>8\)

Vô nghiệm

5(x+y)²+3(x-y)²=8x²+4xy+8y²=4(2x²+xy+2z²)>=5(x+y)²

^{=> \(\sqrt{2x^2+xy+2y^2}\ge\sqrt{\frac{5\left(x+y\right)^2}{4}}\)= \(\frac{\sqrt{5}\left(x+y\right)}{2}\)}

^{Tương tự. Cộng lại là ra nha. Dấu = xảy ra <=> x=y=z=1/3}

x+y+z=0

=>x+y=-z; x+z=-y; y+z=-x

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

=>\(x^2+2xy+y^2=z^2\)

=>\(z^2-xy=x^2+xy+y^2\)

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

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

\(xy+2z^2=xy+2(x^2+2xy+y^2)\)

\(=2x^2+5xy+2y^2=(x+2y)(2x+y)\)

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

=>\(x^2+2xz+z^2=y^2\)

=>\(y^2-xz=x^2+xz+z^2\)

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

\(=4yz-\left(y^2+2yz+z^2\right)=-y^2+2yz-z^2=-\left(y-z\right)^2\)

\(yz+2x^2=yz+2\left(y^2+2yz+z^2\right)\)

\(=2y^2+5yz+2z^2=2y^2+4yz+yz+2z^2\)

=2y(y+2z)+z(y+2z)

=(y+2z)(2y+z)

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

=>\(y^2+2yz+z^2=x^2\)

=>\(x^2-yz=y^2+yz+z^2\)

\(4xz-y^2\) =4xz-(x+z)^2

=4xz-\(x^2-2xz-z^2\)

\(=-x^2+2xz-z^2=-\left(x-z\right)^2\)

\(xz+2y^2=xz+2\left(x+z\right)^2\)

\(=xz+2x^2+4xz_{}+2z^2=2x^2+5xz+2z^2\)

\(=2x^2+4xz+xz+2z^2\)

=2x(x+2z)+z(x+2z)

=(x+2z)(2x+z)

2x+y=x+x+y=x-z

\(2y + z = y + (y + z) = y - x = -(x - y)\)

\(2z + x = z + (z + x) = z - y = -(y - z)\)

\(x + 2y = (x + y) + y = -z + y = y - z\)

\(y + 2z = (y + z) + z = -x + z = z - x\)

\(z + 2x = (z + x) + x = -y + x = x - y\)

Ta có: \(A = \frac{4xy - z^2}{xy + 2z^2} \cdot \frac{4yz - x^2}{yz + 2x^2} \cdot \frac{4zx - y^2}{xz + 2y^2}\)

\(=\frac{-(x - y)^2}{(x + 2y)(2x + y)}\cdot\frac{-(y - z)^2}{(y + 2z)(2y + z)}\cdot\frac{-(z - x)^2}{(z + 2x)(2z + x)}\)

\(=\frac{-\left(x-y\right)^2\cdot\left(y-z\right)^2\cdot\left(z-x\right)^2}{(y-z)(x-z)(z-x)[-(x-y)](x-y)[-(y-z)]}\)

\(=\frac{-(x - y)^2 (y - z)^2 (z - x)^2}{-(x - y)^2 (y - z)^2 (z - x)^2}=1\)