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From the examples in the O.P., I assume that we are dealing with homogeneous quadratic polynomials, that is, with quadratic forms.

Let us state the problem: given a quadratic form $p$ in $\mathbb{R}^n$, determine the truth value of the proposition

(P) there exists $\mathbf{x}\in\mathbb{R}_+^n$ such that $p(\mathbf{x})>0$.

Here, $\mathbb{R}_+$ stands for the set of positive real numbers. One can apply, at least, one of the following tests:

Test 1. Let $A$ be the symmetric matrix $A$ associated to $p$, i.e., $p(\mathbf{x})=\mathbf{x}^TA\mathbf{x}$. Compute the eigenvalues of $A$. If all the eigenvalues are less than or equal to $0$, then (P) is false.

Test 2. Compute the maximum of $p$ on the set $D=[0,1]^n$. Then (P) is true if and only if such a maximum is positive.

Test 1 can be implemented through the eigenvalues() method for matrices. Test 2 can use the minimize_constrained function, already presented in @dan_fulea’s comment. Please note that maximize $p$ is equivalent to minimize $-p$.

Let us apply them to the given polynomials.

Example 1. Let $p=-x^2 - y^2 - z^2 + xy + xz + yz$. We apply Test 1:

sage: var("x,y,z");
sage: A = matrix([[-1,1/2,1/2],[1/2,-1,1/2],[1/2,1/2,-1]])
sage: A.eigenvalues()
[0, -3/2, -3/2]


By Test 1, (P) is false. We could also apply Test 2:

sage: var("x,y,z");
sage: p = -x^2 - y^2 - z^2 + x*y + x*z + y*z
sage: sol = minimize_constrained(-p, [[0,1]]*3, [1,1,1])
sage: print "Maximum is", p(*sol), "attained at", sol
Maximum is 0.0 attained at (1.0, 1.0, 1.0)


Since the maximum is not greater than $0$, by Test 2, (P) is false. The second argument of minimize_constrained is a list of the intervals where $x$, $y$ and $z$ should be, that is, $[0,1]$ for each variable. The last argument is a starting point of the iterative minimization process.

Example 2. Let $q= -x^2 - y^2 - z^2 + (3/2)(xy + xz + yz)$. We first use Test 1:

sage: var("x,y,z");
sage: A = matrix([[-1,3/4,3/4],[3/4,-1,3/4],[3/4,3/4,-1]])
sage: A.eigenvalues()
[1/2, -7/4, -7/4]


Since $A$ has one positive eigenvalue, Test 1 fails. We need to apply Test 2:

sage: var("x,y,z");
sage: q = -x^2 - y^2 - z^2 + (3/2)*(x*y + x*z + y*z)
sage: sol = minimize_constrained(-q,[[0,1]]*3, [1,1,1])
sage: print "Maximum is", q(*sol), "attained at", sol
Maximum is 1.5 attained at (1.0, 1.0, 1.0)


Since the maximum is positive, by Test 2, (P) is true.

Rationale for Test 1. If $A$ does not have a positive eigenvalue, then $A$ is semidefinite negative, and so $p(\mathbf{x})=\mathbf{x}^TA\mathbf{x}\leq 0$ for all $\mathbf{x}\in\mathbb{R}^n$. Consequently, (P) is false.

Rationale for Test 2. Since $p$ is continuous and $D$ is compact, there always exists at least one point $\mathbf{x}_0\in D$ where $p$ attains a maximum value in $D$, i.e. $p(\mathbf{x}_0)\geq p(\mathbf{x})$ for all $\mathbf{x}\in D$. Now, if (P) is true, there exist $\mathbf{x}_1\in \mathbb{R}_+^n$ such that $p(\mathbf{x}_1)>0$. Let $\mathbf{x}_2=\mathbf{x}_1/\lVert\mathbf{x}_1\rVert$, which obviously belongs to $D$. We deduce that $p(\mathbf{x}_0) \geq p(\mathbf{x}_2)=\frac{p(\mathbf{x}_1)}{\lVert\mathbf{x}_1\rVert^2}>0.$ Conversely, assume that $p(\mathbf{x}_0)>0$. If $\mathbf{x}_0$ belongs to the interior of $D$, then (P) holds with $\mathbf{x}=\mathbf{x}_0$. If $\mathbf{x}_0$ is in the boundary of $D$ (so possibly with a null coordinate), by continuity of $p$, there exists a ball $B$ centered at $\mathbf{x}_0$ where the sign of $p$ is that of $p(\mathbf{x}_0)$, that is, $p$ is positive on $B$. Since $B$ contains at least one point $\mathbf{x}_3$ in the interior of $D$, then (P) holds with $\mathbf{x}=\mathbf{x}_3$.

From the examples in the O.P., I assume that we are dealing with homogeneous quadratic polynomials, that is, with quadratic forms.

Let us state the problem: given a quadratic form $p$ in $\mathbb{R}^n$, determine the truth value of the proposition

(P) there exists $\mathbf{x}\in\mathbb{R}_+^n$ such that $p(\mathbf{x})>0$.

Here, $\mathbb{R}_+$ stands for the set of positive real numbers. One can apply, at least, one of the following tests:

Test 1. Let $A$ be the symmetric matrix $A$ associated to $p$, i.e., $p(\mathbf{x})=\mathbf{x}^TA\mathbf{x}$. Compute the eigenvalues of $A$. If all the eigenvalues are less than or equal to $0$, then (P) is false.false. If there exists a positive eigenvalue $\lambda$ with an associated eigenvector $\mathbf{v}\in\mathbb{R}_+^n$, then (P) is true.

Note that this test does not cover all the possibilities, since it only checks sufficient conditions.

Test 2. Compute the maximum of $p$ on the set $D=[0,1]^n$. Then (P) is true if and only if such a maximum is positive.

Test 1 can be implemented through the eigenvalues() method and eigenvectors_right() methods for matrices. Test 2 can use the minimize_constrained function, already presented in @dan_fulea’s comment. answer. Please note that maximize $p$ is equivalent to minimize $-p$.

Let us apply them to the given polynomials.

Example 1. Let $p=-x^2 - y^2 - z^2 + xy + xz + yz$. We apply Test 1:

sage: var("x,y,z");
sage: A = matrix([[-1,1/2,1/2],[1/2,-1,1/2],[1/2,1/2,-1]])
sage: A.eigenvalues()
[0, -3/2, -3/2]


By Test 1, (P) is false. false, since all the eigenvalues are less than or equal to $0$. We could also apply Test 2:

sage: var("x,y,z");
sage: p = -x^2 - y^2 - z^2 + x*y + x*z + y*z
sage: sol = minimize_constrained(-p, [[0,1]]*3, [1,1,1])
[0.1,0.9,0.5])
sage: print "Maximum is", p(*sol), "attained at", sol
Maximum is 0.0 attained at (1.0, 1.0, 1.0)
(0.4999999998931121, 0.5000000001068878, 0.5)


Since the maximum is not greater than $0$, by Test 2, (P) is false. false. The second argument of minimize_constrained is a list of the intervals where $x$, $y$ and $z$ should be, that is, $[0,1]$ for each variable. The last argument is a starting point of the iterative minimization process.process. In this example, if we take a different starting point, the maximum is also $0$, but it can be reached at a different point ($p$ is $0$ on the line $x=y=z$).

Example 2. Let $q= -x^2 - y^2 - z^2 + (3/2)(xy + xz + yz)$. We first use Test 1:

sage: var("x,y,z");
sage: A = matrix([[-1,3/4,3/4],[3/4,-1,3/4],[3/4,3/4,-1]])
sage: A.eigenvalues()
[1/2, -7/4, -7/4]
sage: A.eigenvectors_right()
[(1/2, [(1, 1, 1)], 1), (-7/4, [(1, 0, -1), (0, 1, -1)], 2)]


Since The matrix $A$ has one positive eigenvalue, Test 1 fails. We need $\lambda=1/2$, with an associated eigenvector $\mathbf{v}=(1,1,1)$ belonging to $\mathbb{R}_+^n$. Hence, (P) is true. Let us now apply Test 2:

sage: var("x,y,z");
sage: q = -x^2 - y^2 - z^2 + (3/2)*(x*y + x*z + y*z)
sage: sol = minimize_constrained(-q,[[0,1]]*3, [1,1,1])
[0.1,0.9,0.5])
sage: print "Maximum is", q(*sol), "attained at", sol
Maximum is 1.5 attained at (1.0, 1.0, 1.0)


Since the maximum is positive, by Test 2, (P) is true.

Rationale for Test 1. If $A$ does not have a positive eigenvalue, then $A$ is semidefinite negative, and so $p(\mathbf{x})=\mathbf{x}^TA\mathbf{x}\leq 0$ for all $\mathbf{x}\in\mathbb{R}^n$. Consequently, (P) is false.false. Likewise, if there exists a positive eigenvalue $\lambda$ with an associated eigenvector $\mathbf{v}\in\mathbb{R}_+^n$, then $p(\mathbf{v})=\mathbf{v}^TA\mathbf{v}=\lambda\mathbf{v}^T\mathbf{v}=\lambda\lVert\mathbf{v}\rVert^2>0$. Hence (P) is true.

Rationale for Test 2. Since $p$ is continuous and $D$ is compact, there always exists at least one point $\mathbf{x}_0\in D$ where $p$ attains a maximum value in $D$, i.e. $p(\mathbf{x}_0)\geq p(\mathbf{x})$ for all $\mathbf{x}\in D$. Now, if (P) is true, there exist $\mathbf{x}_1\in \mathbb{R}_+^n$ such that $p(\mathbf{x}_1)>0$. Let $\mathbf{x}_2=\mathbf{x}_1/\lVert\mathbf{x}_1\rVert$, which obviously belongs to $D$. We deduce that $p(\mathbf{x}_0) \geq p(\mathbf{x}_2)=\frac{p(\mathbf{x}_1)}{\lVert\mathbf{x}_1\rVert^2}>0.$ Conversely, assume that $p(\mathbf{x}_0)>0$. If $\mathbf{x}_0$ belongs to the interior of $D$, then (P) holds with $\mathbf{x}=\mathbf{x}_0$. If $\mathbf{x}_0$ is in the boundary of $D$ (so possibly with a null coordinate), by continuity of $p$, there exists a ball $B$ centered at $\mathbf{x}_0$ where the sign of $p$ is that of $p(\mathbf{x}_0)$, that is, $p$ is positive on $B$. Since $B$ contains at least one point $\mathbf{x}_3$ in the interior of $D$, then (P) holds with $\mathbf{x}=\mathbf{x}_3$.