Find the saddle point of $F(x_1,x_2,x_3,y_1,y_2,y_3)=(x_1-2x_2+x_3)y_1+(2x_1-2x_3)y_2+(-x_1+x_2)y_3$ Finding the saddle points of $F(x_1,x_2,x_3,y_1,y_2,y_3)=(x_1-2x_2+x_3)y_1+(2x_1-2x_3)y_2$+$(-x_1+x_2)y_3$ subject to the constraints $x_1+x_2+x_3=1, y_1+y_2+y_3=1$. Show that the saddle point is $x=(\frac{1/3}{1/3},\frac{1/3}{1/3},\frac{1/3}{1/3}),y=(\frac{2}{7},\frac{1}{7},\frac{4}{7})$.
I know how to find the saddle point of a function with two variable by using $\Delta=(f_{12})^2-f_{11}f_{22}$, then when $\Delta>0$ the point is a saddle point. 
But for this question, by using the constraints, we can reduce the variables from 6 to 4, but still can not use the $\Delta$ formula, and this question requires not to use lagrange multiplier. Thanks.
 A: At a saddle point, all of the partial derivatives are 0. This leads to a system of equations:
\begin{align}
x_1 -2x_2 +x_3=0,\tag{1}\\
2x_1 -2x_3=0,\tag{2}\\
x_2-x_1=0,\tag{3}\\
y_1+2y_2-y_3=0, \tag{4}\\
-2y_1 +y_3=0,\tag{5}\\
y_1-2y_2=0. \tag{6}
\end{align}
Adding 1 to both sides of equation (4) leads to
\begin{equation}
2 y_1 = 1-3 y_2. \tag{7}
\end{equation}
Solving for $y_1$ from equation (6) and substituting in (7) leads to $y_2= \frac{1}{7}$ and consequently $y_1=2y_2=\frac{2}{7}$ and $y_3=2y_1=\frac{4}{7}$.
Similarly, we can subtract 1 from both sides of equation (1) and use the constraint for x, which leads to $x_2=\frac{1}{3}$ and consequently from (3), $x_1=x_2=\frac{1}{3}$. Finally, from (3), we have $x_3=x_1=\frac{1}{3}$.
EDIT:
Computing the Hessian matrix yields:
\begin{equation}
H = 
\begin{pmatrix}
0 & 0 & 0 & 1 & 2 & -1 \\
0 & 0 & 0 & -2 & 0 & 1\\
0 & 0 & 0 & 1 & 0 & 0\\
1 & -2 & 1 & 0 & 0 & 0\\
2 & 0 & 0 & 0 & 0 & 0\\
-1 & 1 & 0 & 0 & 0 & 0
\end{pmatrix}.
\end{equation}
Diagonalizing gives us the characteristic polynomial
\begin{equation}
\lambda^6-12\lambda^4 +27\lambda^2 -4=0.
\end{equation}
This polynomial has a $\lambda \rightarrow -\lambda$ symmetry. So if we can find one non-zero real solution, we automatically show that the Hessian is indefinite (There are both positive and negative solutions). Therefore the critical point above is indeed a saddle point.
EDIT 2:
In case you find it difficult to actually find the roots of the polynomial, substitute $\rho=\lambda^2$, this gives
\begin{equation}
\rho^3-12\rho^2 +27\rho -4=0.
\end{equation}
Now, using the fact that complex roots come in pairs and that we can see that zero is not a solution, we can deduce that there is at least one real solution of the cubic $\rho_0$ and therefore two opposing sign solutions $\lambda_{+/-} = \pm \sqrt{|\rho_0|}$.
