The definition of centre of mass on Wikipedia is given as
This is the point to which a force may be applied to cause a linear acceleration without an angular acceleration.
How can I prove that such a point is the weighted average of the radius vectors of all discrete masses mathematically?
Also see this question.
Because force is the time derivative of momentum, and momentum is linked to the motion of the center of mass.
If you consider a rigid body as a collection of particles glued together and their position split into the position of the center of mass $\boldsymbol{r}_{\rm COM}$ plus some other relative position $\boldsymbol{d}_i$, then
$$ \boldsymbol{r}_i = \boldsymbol{r}_{\rm COM} + \boldsymbol{d}_i $$
and by taking the weighted average of the positions
$$\require{cancel} \sum_i m_i \boldsymbol{r}_i = \left( \sum_i m_i \right) \boldsymbol{r}_{\rm COM} + \cancel{ \sum_{i} m_i \boldsymbol{d}_i } $$
which means that the center of mass is the point which the weighted average relative position is zero $\sum_i m_i \boldsymbol{d}_i = 0$.
Now consider the motion of each particle as the velocity of the center of mass, and a rotation about the center of mass
$$ \boldsymbol{v}_i = \boldsymbol{v}_{\rm COM} + \boldsymbol{\omega} \times \boldsymbol{d}_i $$
Use the above to consider linear and angular momentum
Linear Momentum
$$\boldsymbol{p} = \sum_i m_i \boldsymbol{v}_i = \left( \sum_i m_i \right) \boldsymbol{v}_{\rm COM} + \boldsymbol{\omega} \times \left( \cancel{ \sum_i m_i \boldsymbol{d}_i }\right) = m\, \boldsymbol{v}_{\rm COM} $$
Angular Momentum about center of mass
$$ \begin{aligned} \boldsymbol{L}_{\rm COM} & = \sum_i \boldsymbol{d}_i \times (m_i \boldsymbol{v}_i) \\ &= \left( \cancel{ \sum_i m_i \boldsymbol{d}_i} \right) \times \boldsymbol{v}_{\rm COM} + \sum_i \boldsymbol{d}_i \times m_i ( \boldsymbol{\omega} \times \boldsymbol{d}_i) \\ &= \mathbf{I}_{\rm COM}\; \boldsymbol{\omega} \end{aligned}$$
The last part of the puzzle is equating net force $\boldsymbol{F}$ to the rate of change of linear momentum and net torque about the center of mass $\boldsymbol{\tau}_{\rm COM}$ to the rate of change of angular momentum.
The equations below are the standard equations of motion for a rigid body.
$$ \boxed{ \begin{aligned} \boldsymbol{F} &= \tfrac{\rm d}{{\rm d}t} \boldsymbol{p} = m\,\boldsymbol{a}_{\rm COM} \\ \boldsymbol{\tau}_{\rm COM} & = \tfrac{\rm d}{{\rm d}t} \boldsymbol{L}_{\rm COM} = \mathbf{I}_{\rm COM} \boldsymbol{\alpha} + \boldsymbol{\omega} \times \boldsymbol{L}_{\rm COM}\end{aligned} }$$
So consider a force $\boldsymbol{F}$ applied away from the center of mass, which causes a net torque $\boldsymbol{\tau}_{\rm COM} = \boldsymbol{d} \times \boldsymbol{F} \neq 0 $ to a body initially at rest. This means that $\boldsymbol{\alpha} \neq 0$ causing rotational acceleration.
In summary, although a force applied on a body with always accelerate the center of mass, only a force through the center of mass causes no net torque, which would keep the body for accelerating rotationally.
Let $\vec r_1$ denote the position of the centre of mass of an object of mass $m$, given by the formula below. $$\vec r_1 = \frac{1}{m}\int \rho \vec r^\prime \mathrm{d^3} \vec r^\prime$$
If an object is not rotating, then all of its points must have the same acceleration. Therefore, if a single force applied to the object does not cause rotation, it must be uniformly distributed over the mass. Let $\vec r_2$ denote the point at which a force can be applied to not cause rotation (the centre of mass using the definition you gave).
Consider the torque resulting from a force $\vec F$ applied at $\vec r_2$. Since the force is applied at $\vec r_2$, it is uniformly distributed over every infinitesimal piece of mass $\mathrm{d} m = \rho \mathrm{d^3} \vec r^\prime$. The torque about $\vec r_2$ resulting from this is given below. Note that $\vec r_1$ from above appears in this formula.
$$\vec\tau = \frac{1}{m}\int (\vec r^\prime - \vec r_2) \times \vec F \rho\mathrm{d^3} \vec r^\prime = \frac{1}{m}\left(\int (\vec r^\prime - \vec r_2) \rho\mathrm{d^3} \vec r^\prime \right) \times \vec F = (\vec r_1 - \vec r_2) \times \vec F$$
Since this setup does not cause rotation, the torque must be zero, from which we can conclude that $$\vec r_1 = \vec r_2$$
