I'm looking at the proof of lemma 2 on page 6 of https://cowles.yale.edu/sites/default/files/files/pub/d05/d0549.pdf
Which claims that the LCP $$ Mx+q \geq 0 \\ x \geq 0 \\ x^{T}(Mx+q) = 0 $$ has a solution if and only if the system $$ x^{T}(Mx+q) < 2^{-3L} \\ Mx+q>-2^{-4L}e \\ x > -2^{-4L}e \\ Mx+q < 2^{L}e \\ x < 2^{L}e $$ has a solution; where $q, x, e \in \mathbb{R}^n$, $e$ a vector of ones, $M$ a positive semi-definite $n$ by $n$ matrix, $q$ and $M$ have integer entries ($q_j$ and $M_{i,j}$), and $$L = \sum_i \sum_j \log_2(|M_{i,j}|+1) + \sum_j \log_2(|q_j|+1) + \log_2(n)+1$$ Now the proof begins by assuming $x^0$ is a solution to the above system. Then $$x^0_i(M_ix^0_i+q_i) > - 2^{-3L}$$ where $M_i$ is the $i$th column of $M$, which I follow. The author then claims that this together with the quadratic inequality ($x^{T}(Mx+q) < 2^{-3L}$) implies that $$x^0_i(M_ix^0_i+q_i) < 2^{-3L} + \sum_{j=1}^{n} x^0_j(M_jx^0_j+q_j)$$ which I do not follow. Can't we have for each $j<n$ $x^0_j(M_jx^0_j+q_j)$ arbitrarily close to $-2^{-3L}$ then for $i=n$ this inequality would imply $$0 < -(n-2)2^{-3L} + \epsilon$$ for arbitrarily small $\epsilon$? And for $n \geq 3$ this would be false. The author does not claim that the $x^0$ solution is composed of rational numbers or anything so I'm quite confused. What obvious thing am I missing? Any help is very much appreciated. Thank you for your time and consideration.
Alright after looking at the proof again after some sleep I think I've got the answer.
For each $i \in \{1, \dots, n\}$ let $$y_i = x^0_i(M_ix^0_i+q_i) > -2^{-3L}.$$ With this, it follows from $$x(Mx+q)<2^{-3L}$$ that for every $i \in \{1, \dots, n\}$ $$y_i < n2^{-3L}.$$ Taking this new inequality together with the definition of $L$ and the fact that $$\forall i \in \{1,\dots,n\} (y_i > -2^{-3L}) \implies \forall i \in \{1,\dots,n\} \left ( -(n-1)2^{-3L} < \sum_{j \in \{1,\dots,n\} \setminus \{i\}}y_j \right )$$ it follows that for each $i \in \{1, \dots, n\}$ \begin{align} y_i &< (n-1)2^{-3L} + \sum_{j=1}^{n}y_j \\ &< n2^{-3L} \\ &< 2^{-2L}. \end{align}