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This is a hard problem (maximizing the product is a bit better one, as sometimes one can take $\log$ of the objective function, and it becomes concave...). Your best shot might be to use the sum of sq …

If you like, you might look at cones of sums of squares of polynomials (cones of PSD matrices are the same thing as cones of sums of squares of linear polynomials). This is the starting point of the m …

Let us simplify your constraints. Namely, set $x=r\sin\phi$, $y=r\cos\phi$. Then it simplifies to $u\sin^2\phi + v\cos^2\phi \geq \sin\phi \cos\phi$, for all $0\leq \phi\leq 2\pi.$ Then divide both si …

In theory, there is a polynomial-time algorithm (follows from results of my joint paper with Dima Grigoriev) when $m$ is fixed, but this is not a practical one, and moreover I see from comments above …

To answer on methods applicable here (and elaborate on comments I made). The most promising is to use a surprisingly little-known theorem that says that the discriminant $D$ of a symmetric $n\times n$ …

No, this is not true (unless P=NP). There are examples of convex optimization problems which are NP-hard. Several NP-hard combinatorial optimization problems can be encoded as convex optimization prob …

This is the well-known problem: http://en.wikipedia.org/wiki/Quadratic_assignment_problem In general there is no efficient algorithm known for this problem. A much more famous Travelling Salesman Pro …

This is a hard question, in view of Motzkin-Straus theorem (cf. e.g. 5.2.4 here), which in particular says that the local minima of $f(x):=x^\top (I+A) x$, where $A$ is the adjacency matrix of a graph …

It's a long comment, not an answer. Already the well-known SDR cone $\Sigma_{n,d}$ of sums of squares of homogeneous $n$-variate degreed $d$ polynomials has an interesting dual (also SDR), described …

If $g_j$ is SOS, i.e. $g_j=\sum_k h_k^2$, then $g_j(x)\leq 0$ iff $g_j(x)=0$, i.e. $h_1(x)=h_2(x)=\dots =0$. So this is a general case, although with equality constraints only. A more interesting que …