Any $(x, y, z)$ can satisfy the $5x^2+2y^2+6z^2-6xy-2xz+2yz<0$?

Question

Please tell me whether there any $(x, y, z)$

which can satisfy the $5x^2+2y^2+6z^2-6xy-2xz+2yz<0$ ?

No process or just solve it by calculator are both fine.

Thank you.

Answer 1

We have \begin{align} &5x^2 + 2y^2 + 6z^2 - 6xy - 2xz + 2yz\\ = \ & 2y^2 + (-6x + 2z)y + 5x^2 - 2xz + 6z^2\\ = \ & 2\left(y + \frac{-3x+z}{2}\right)^2 - 2 \left(\frac{-3x+z}{2}\right)^2 + 5x^2 - 2xz + 6z^2\\ = \ & 2\left(y + \frac{-3x+z}{2}\right)^2 + \frac{1}{2}x^2 + xz + \frac{11}{2}z^2 \\ = \ & 2\left(y + \frac{-3x+z}{2}\right)^2 + \frac{1}{2}(x + z)^2 - \frac{1}{2}z^2 + \frac{11}{2}z^2\\ = \ & 2\left(y + \frac{-3x+z}{2}\right)^2 + \frac{1}{2}(x+z)^2 + 5z^2. \end{align} Thus, $5x^2 + 2y^2 + 6z^2 - 6xy - 2xz + 2yz$ is never negative.

Answer 2

The expression $$5x^2+2y^2+6z^2-6xy-2xz+2yz$$ may be written as $$3(x-y)^2+(x-z)^2+(y+z)^2+x^2-2y^2+4z^2$$ Notice that the only term with a negative sign is $-2y^2$, all others are positive. Setting $x=z=0$ we have $$3(x-y)^2+(x-z)^2+(y+z)^2+x^2-2y^2+4z^2\geq2y^2\geq 0$$

Hence, the expression is never negative.

Answer 3

Your expression is the quadratic form $$\begin{bmatrix}x&y&z\end{bmatrix}\begin{bmatrix}5&-3&-1\\-3&2&1\\-1&1&6\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}.$$ Let's call the 3x3 matrix above $A.$ Since, in the matrix $A$, the upper left element, the upper left 2x2 determinant and the determinant of the whole matrix are all positive, the quadratic expression is positive, except when $x,y,z$ are all 0. Thus the given expression is never negative.

Answer 4

Consider that you search for the extremum of $$F=5x^2+2y^2+6z^2-6xy-2xz+2yz$$ $$\frac{\partial F}{\partial x}=10 x-6 y-2 z=0 \qquad \frac{\partial F}{\partial y}=-6 x+4 y+2 z=0 \qquad \frac{\partial F}{\partial z}=-2 x+2 y+12 z=0$$

The only solution is $x=y=z=0$ so the minimum value of $F$ is $0$.

Then $F$ is non-negative for any $(x,y,z)$.