Boyd & Vandenberghe, problem 4.55 — how to show that solution is Pareto optimal? In problem 4.55 of Boyd & Vandenberghe's Convex Optimization, the authors ask the following.

Show that in a multicriterion optimization problem, a unique solution of the scalar optimization problem
$$ \min. \max._{i =1,2\cdots q}F_i (x) 
$$
$$\text{s.t. } f_i(x)\leq 0
$$
$$
h_i(x)=0
$$ is Pareto optimal.

I know that, in a multicriterion optimization problem a solution is pareto optimal point if we can not find a better point. I assume that $x^*$ is the solution of the above scalar optimization problem. Now I have to show that for every feasible $y\neq x^*$ we have $$[F_1(x^*),~F_2(x^*), \cdots F_q(x^*)]\preceq [F_1(y),~F_2(y), \cdots F_q(y)].$$ I think the only other information that I have is that one of the $F_i(x^*)$ is greater than all of the rest of $F_j(x^*)'s$ for $j\neq i$. How to solve this problem? Thanks in advance.

Assume $x^*$ is not Pareto optimal then wlog there exists $y \neq x^*$ such that $F_1(y) < F_1(x^*)$ and $F_i(y) \leq F_i(x^*)$ for all $i =2\cdots q$. Now, this implies
$$ \max_{i =1,2\cdots q}F_i (y) \leq \max_{i =1,2\cdots q}F_i (x^*) = \min_{x} \max_{i =1,2\cdots q}F_i (x),$$
by definition of $x^*$. Therefore, $y$ is also a minimum to the original scalar optimization problem. By assumption, this minimum is unique, therefore $y = x^*$, a contradiction. Therefore, $x^*$ is Pareto optimal.
