I'm trying to find the largest inscribed ball in a polyhedron $P$.
$P = \{ x \in \mathbb{R}^n : P_x*x \leq P_c \}$, $P_x$ is a matrix $n_P*n$ and $P_c$ is a vector with dimension $n_P$.
On my book it says that the center and radius of the Chebyshev Ball (largest ball) can be easily found by solving the following linear optimization problem
$$ \begin{align*} \underset{x_c, R}{\text{max}} &~ R \tag{4.18} \\ \text{subj. to} &~ P^x_i x_c +R \Vert P^x_i \Vert_2 \leq P_i^c, \quad i=1,\ldots, n_P\\ &~ R \geq 0 \end{align*} $$
I would like to know how to get to this formulation. Particularly on my book it says that the proof is the following: the problem can be written as
$$ \begin{align*} \underset{x_c, R}{\text{max}} &~ R \tag{4.19}\\ \text{subj. to} &~ P^x_i (x_c + v) \leq P_i^c, \text{ $\forall v$ such that $\Vert v \Vert_2 \leq R, i=1,\ldots, n_P$}\\ &~ R \geq 0 \end{align*} $$
and considering I-th constrains
$$ P^x_i (x_c + v ) \leq P_i^c, \forall v \text{ such that } \Vert v \Vert_2 \leq R. $$
this can be written as
$$ P^x_i x_c \leq P_i^c - P^x_i v, \forall v \text{ such that } \Vert > v \Vert_2 \leq R. \tag{4.20}$$
And now (the following) is the step I really don't get, so I hope you can help with.
Constraint (4.20) is satisfied $\forall v$ such that $\Vert v \Vert_2 \leq R$ if and only if it is satisfied at $v = \frac{P_i^{x'}}{ \Vert P_i^x \Vert_2}R$. Therefore we can rewrite the optimization problem (4.19) as the linear program (4.18).
Why is the constrain satisfied for every $v$ if and only if the constrains are satisfied for $v = \frac{P_i^{x'}}{ \Vert P_i^x \Vert_2}R$ ?
The forward direction is trivial; if $\forall v \text{ such that } ||v||_2 \leq R$, then the constraint in (4.20) should hold in particular for $v = \frac{{P_i^{x}}^{'}}{||P_i^x||_2}R$, as it has magnitude $R$.
To show sufficiency, we let $v = \frac{{P_i^{x}}^{'}}{||P_i^x||_2}R$, and consider arbitrary $w \text{ such that } ||w||_2 \leq R$. Note that by Cauchy-Schwarz, $$\langle {P_i^x}^{'}, w \rangle \leq ||P_i^x||_2 \cdot ||w||_2 \leq ||P_i^x||_2 \cdot R$$
Examining $v$, we can see that $P_i^xv=\frac{\langle{P_i^x}^{'}, {P_i^x}^{'} \rangle}{||P_i^x||_2}R= ||P_i^x||_2\cdot R$. We thus conclude that $$\langle {P_i^x}^{'}, w \rangle \leq P_i^x v$$
$$\implies P_i^xw \leq P_i^xv$$
Negating both sides and adding $P_i^c$ yields $$P_i^c-P_i^xw\geq P_i^c-P_i^xv \geq P_i^xx_c. $$