# Exercise: Let P &#x2208;<!-- ∈ --> <mi mathvariant="double-struck">R V </msup>

Exercise: Let $P\in {\mathbb{R}}^{V}$ be defined by the inequalities

node set $V=\left\{1,2,3,4,5,6,7,8,9\right\}$
Starting from the system $\left(1\right)-\left(2\right)$, give the cutting-plane proof of the inequality
${x}_{1}+{x}_{2}+{x}_{6}\ge 2$
What I've tried: I know that I need to show that there exists a nonnegative combination from the inequalities $\left(1\right)$ and $\left(2\right)$ such that ${x}_{1}+{x}_{2}+{x}_{3}\ge 3$ holds. I don't know how unfortunately.
Question: How do I solve this exercise?
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Krish Finley
If we put ${x}_{1}={x}_{2}={x}_{6}=1/2$ and all other ${x}_{i}$’s equal 1 then all the defining inequalities will be satisfied but ${x}_{1}+{x}_{2}+{x}_{6}=3/2<2$.
But the claim holds provided we require all xi are integer. Indeed, if ${x}_{1}+{x}_{2}+{x}_{6}<2$ then al least two of these ${x}_{i}$ (say, ${x}_{u}$ and ${x}_{v}$) are zero. Then $uv\in E$, but ${x}_{u}+{x}_{v}=0$, a contradiction to (2).