Conditions to make conic section positve in the region $0 \leq x+y \leq 1$ Suppose $0 \leq x+y \leq 1$ and let $a,b,c \in [0,1]$ be fixed. I've been stuck with this inequality:
$$ -2\,{b}^{2}{x}^{2}+ \left( -4\,{b}^{2}+4 \right) xy+ \left( -4\,ab-2\,
c-2 \right) x-2\,{b}^{2}{y}^{2}+ \left( -4\,ab+2\,c-2 \right) y-2\,{a}
^{2}-{c}^{2}+1 \ge 0 $$
for a while now. Basically I am trying to find conditions on $a,b,c$ so that there exists some $x,y$ in the given region that makes this inequality true. Clearly the function on the left is some sort of conic in the $x,y$-plane, however, I am not sure what it would look like for a fixed $a,b,c$. Is there a straightforward way to find conditions on the $a,b,c$ so that this conic is positive in the given region?

After a bunch of completing-the-square, we can write the polynomial as $$\begin{bmatrix}x-x_0 & y-y_0\end{bmatrix} \begin{bmatrix}-2b^2 & 2-2b^2 \\ 2-2b^2 & -2b^2\end{bmatrix}\begin{bmatrix}x-x_0 \\ y - y_0\end{bmatrix} + z_0$$
where
$$x_0 = \frac{1-c+2b(a+bc)}{2-4b^2}, \qquad y_0 = \frac{1+c+2b(a-bc)}{2-4b^2}, \qquad z_0 = \frac{2(a+b)^2}{2b^2-1}.$$
The matrix $$\begin{bmatrix}-2b^2 & 2-2b^2 \\ 2-2b^2 & -2b^2\end{bmatrix}$$ has one eigenvalue of $-2$ with eigenvector $(1,-1)$, and one eigenvalue of $2-4b^2$ with eigenvector $(1,1)$. So under the substitution arising from $$\begin{bmatrix}x-x_0 \\ y-y_0 \end{bmatrix} = u \begin{bmatrix} 1 \\ -1\end{bmatrix} + v \begin{bmatrix}1 \\ 1\end{bmatrix}$$ we can rewrite the polynomial as
$$-4u^2 + (4 - 8b^2) v^2 + z_0.$$
The constraint that $0 \le x + y \le 1$ turns into $0 \le 2v + x_0 + y_0 \le 1$, with no restriction on $u$. Since $u$ contributes a nonpositive term $-4u^2$, we should set $u=0$, which is equivalent to saying that $x - x_0 = y - y_0$, or that $y = x+(y_0 - x_0) = x+c$.
Now the polynomial is a quadratic in $x$. Relevantly, the coefficient of $x^2$ is $4-8b^2$. So to maximize the quadratic, we can do one of two things.


*

*If $b \le \frac1{\sqrt2}$, then the coefficient of $x^2$ is nonnegative, so the function is maximized at a boundary point. Either $x+y = 2x+c = 0$, or $x+y = 2x+c=1$. At the first point $(x,y) = (-c/2, c/2)$, the polynomial is $1 - 2a^2$, which is sometimes positive. At the second point $(x,y) = ((1-c)/2, (1+c)/2)$, the polynomial is $-2(a+b)^2$, which is always negative except I guess when $a=b=0$.

*If $ b > \frac1{\sqrt2}$, then the coefficient of $x^2$ is negative, so the function is maximized at the vertex of the parabola. Do some algebra and you find that the vertex of the parabola is at $x = \frac{-1-2ab+c-2b^2c}{4b^2-2}$, where $x+y = 2x + c = \frac{1+2ab}{1-2b^2}$ is always negative. That's outside our allowed range, so once again we want to take one of the boundary points above.


As a result, the best point to use is always $(x,y) = (-c/2, c/2)$. At this point, the polynomial is equal to $1 - 2a^2$, which is nonnegative if $a \le \frac1{\sqrt2}$. 
