I am to prove that the constraint $x_1x_2=x_3x_4$ (where $x=(x_1, x_2, x_3, x_4)\in \mathbb{R}^n$ and $x_k\geq 0$ for $k=1,2,3,4$) cannot be written as a second-order cone constraint. In other words, I am to prove the following statement:
There do not exist $A\in \mathbb{R}^{m\times n}, b\in \mathbb{R}^{m}, c\in \mathbb{R}^{n}, d\in \mathbb{R}$ such that [($x_1x_2=x_3x_4$ and $x_k\geq 0$ for $k=1,2,3,4$) implies $\|Ax+b\|\leq c^T x+d$ (where $x=(x_1, x_2, x_3, x_4)\in \mathbb{R}^n$ and $\|\cdot\|$ is the standard Euclidean norm)].
My attempt to start a proof by contradiction is summarized in the following steps:
- Assume the negation of the above statement to be true
- Rewrite $\|Ax+b\|$ as $\sqrt{(Ax+b)^T(Ax+b)}=\sqrt{x^TA^TAx+b^TAx+x^TA^Tb+b^Tb}=\sqrt{\sum_{i,j}(A_i^TA_j x_ix_j) +b^TAx+x^TA^Tb+b^Tb}$
- Use the result of 2. to rewrite $\|Ax+b\|\leq c^T x+d$ as $\sqrt{\sum_{i,j}(A_i^TA_j x_ix_j) +b^TAx+x^TA^Tb+b^Tb}\leq c^Tx+d$ where $A_k$ is the $k$th column of $A$ and indices $i$ and $j$ range from $1$ to $m$ and $1$ to $n$, respectively
- Replace all occurrences of $x_1x_2$ with $x_3x_4$ in the second inequality of 3.
Is this approach correct? If so, how can I argue towards a contradiction (I tried to reach a contradiction that one or more of the entries of $x$ are negative by using the assumption that the entries of $x$ are nonnegative as well as the result of 4., but I was unable to do so)? If not, what other approaches could be considered? Is there a standard way to formally prove that a constraint from an optimization problem is not a second-order cone constraint?
Note: The optimization problem in consideration has decision variable vector $x$