If $x \succsim_i x_i^*$ then $p \cdot x_i\ge w_i.$ (MWG 16.C.2)

Question

(MWG 16.C.2) Suppose that the preference relation $\succsim_i$ is locally nonsatiated and that $x_i^*$ is maximal for $\succsim_i$ in set $\{x_i \in X_i: p \cdot x_i \le w_i\}$. Prove that the following property holds: "If $x \succsim_i x_i^*$ then $p \cdot x_i\ge w_i.$"

Locally nonsatiation means that for every $x_i \in X_i$ and $\epsilon >0$, there exists $x_i' \in X_i$ such that $||x_i - x_i'|| < \epsilon$ and $x_i' \succ x_i$. I think that I need to prove that if $x_i \sim_i x^*_i $ for $x_i \not= x_i^*$, then it is possible that $p\cdot x_i = w_i$ because I already know that if $x_i \succ x_i^*$, then $p\cdot x_i > w_i$.

I am stuck and cannot proceed. Can anyone give me some hint for this question?

Answer 1

Hint: Prove by contradiction. That is, suppose $x_i\succsim_ix_i^*$ but $p\cdot x_i<w_i$, show that this would lead to a contradiction of $x_i^*$ being maximal for the feasible set.